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Only one option is correct. 1. One radian ≈ (a) 57 17 '44.8"° (b) 58 37 '22.4"° (c) 59 27 '33.6"° (d) 56 57 '11.2"° 2. The radius of the circle whose arc of length 15 cm makes an angle of 34 radian at the centre, is (a) 10cm (b) 111 cm4 (c) 20cm (d) 122 cm2 3. The length of an are of a circle of radius 28 cm, that subtends an angle of 45° at the centre, is (a) 12 cm (b) 16 cm (c) 22 cm (d) 24 cm 4. The angle subtended at the centre of a circle of diameter 50 cm by an arc of length 11 cm is (a) 20 15'° (b) 22 48 '° (c) 25 12 '° (d) 27 24 '° 5. The acute angle in radians between the minute and the hour hand of a clock when the time is 4 : 20 is (a) 18π (b) 9π (c) 6π (d) 3π 6. The angles of a triangle are in A.P. , the smallest angle being 4π , the largest angle is (a) 3π (b) 23π ( c) 49π (d) 512π 7. A circular wire of radius 15 cm is cut and bent so as to lie along the circumference of a loop of radius 120 cm. The angle subtended by it at the centre is (a) 30° (b) 45° (c) 60° (d) None of these 8. If the interior angles of a polygon are in A.P. with common difference 5° and the smallest angle 120° , then the number of sides of the polygon is (a) 7 (b) 9 (c) 7 or 15 (d) 9 or 16 9. A horse is tied to a post by a rope. If it moves along a circular path, keeping the rope tight and describes 132 m when it has traced out an angle of 108° at the centre, then the length of the rope is (a) 54 m (b) 66 m (c) 70 m (d) 81 m 10. Find the measures of C∠ of ABC∆ in degrees and radians, if 42m A∠ = ° and ( )/ 6 c

m B π∠ = (a) c35π

(b) c5π

(c) c32π

(d) c53π

11. A semicircle is divided into two sectors whose angles are in the ratio 4 : 5. Find the ratio of their areas (a) 5 :1 (b) 4 : 5 (c) 5 : 4 (d) 3 : 4 12. Two concentric circles with the centre O have radii 18 cm and 36 cm. A line OX cuts the circles in

A and A′ respectively. This line is rotated through an angle of 80° and, in its final position, it cuts the circles in B and B′ respectively. Find the perimeter and the area of the region between the circles and lines (a) 270 cmπ (b) 2288 cmπ (c) 2216 cmπ (d) 2210 cmπ 13. Find the measure of the angle between the hour-hand and the minute-hand of a clock at twenty minutes past two.

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(a) 50° (b) 60° (c) 54° (d) 65° 14. The angle between the hands of a clock when the time is 3.20 p.m. is (a) 18π (b) 9π (c) 6π (d) 8π 15. If the angles of a triangle are in the ratio 1: 2 : 3 , then the smallest angle in radian is (a) 3π (b) 6π (c) 2π (d) 9π 16. A circular wire of radius 5 cm is cut and bent again into an arc of a circle of radius 10 cm. The angle subtended by the arc at the centre in radian is (a) 2π (b) 3π (c) 6π (d) π 17. A circular wire of diameter 40 cm is cut and placed along the circumference of a circle of radius 1 metre, the angle subtended by the wire at the centre of the circle is equal to (a) 4π (b) 3π (c) 25π (d) 6π 18. The central angle of a sector of circle of area 9π sq. cm is 60° , the perimeter of the sector is (a) π (b) 3 π+ (c) 6 π+ (d) 6 19. The angles of a triangle are in A.P., and the greatest angle is double the least then the angles in degrees are (a) 30 , 60 , 90° ° ° (b) 40 , 60 , 80° ° ° (c) 20 , 40 , 80° ° ° (d) 30 , 60 , 80° ° ° 20. A wheel makes 120 revolutions in one minute. Then the angle in radian covered by the wheel in 2 seconds is (a) 4π (b) 6π (c) 8π (d) 2π 21. The ratio of the number of sides of two regular polygons is 3 : 4 and the difference between their angles is 15 .° The number of sides of the polygon having larger angle is (a) 6 (b) 8 (c) 10 (d) 12 22. If cos cos 1 , 0, 0,3 4 5 2 2A B

A Bπ π

= = − < < − < < then the value of 2sin 4sinA B+ is equal to (a) 4− (b) 0 (c) 2 (d) 4 23. If tan ,p

qθ = what is the value of sin cos ?sin cosθ θ

θ θ+−

(a) 1 (b) p q

p q

+−

(c) p q

p q

−+

(d) p q

q p+

24. cos9 sin 9cos9 sin 9° + °°− °

is equal to (a) tan 26° (b) tan 36° (c) tan 54° (d) tan 81° 25. If cos sin 1and 0θ θ θ π− = < ≤ then θ is equal to (a) 6π (b) 3π (c) π (d) None 26. The value of cos1 cos 2 .....cos100° ⋅ ° ° is



(a) 1− (b) 0 (c) 1 (d) None of these 27. If 2 2cos sec ,p θ θ= + then (a) 1p < (b) 0 1p< ≤ (c) 1 2p≤ ≤ (d) 2p ≥ 28. The value of ( ) ( ) ( )cos 270 cos 90 sin 270 cosθ θ θ θ°+ ° − − °− is (a) 1− (b) 0 (c) 12 (d) 1 29. The value of the expression cos1 .cos 2 ..cos179° ° ° equals (a) 1− (b) 0 (c) 12 (d) 1 30. tan 1 sec1 sec tanA A

A A

++

+ is equal to

(a) 2cosec A (b) 2sec A (c) 2sin A (d) 2cos A 31. 4 4cos sinθ θ− is equal to (a) 21 2sin 2θ −

(b) 21 2sin 2θ +

(c) 22cos 1θ − (d) 21 2cos θ+

32. Evaluate 52sin sin12 12π π

(a) 14− (b) 32− (c) 1 (d) 12 33. If α is a root of 225cos 5cos 12 0, 2πθ θ α π+ − = < < , then sin 2α is equal to : (a) 2425− (b) 1318− (c) 1318 (d) 2425 34. If ( ) ( ) 1sin cos ,2x y x y− = + = then values of andx y [ ]( )Given, , 0, 180x y∈ ° are given by (a) 180 , 135x y= ° = ° (b) 165 , 15x y= ° = ° (c) 45 , 135x y= ° = ° (d) 45 , 15x y= ° = ° 35. If θ lies in the second quadrant, then the value of 1 sin 1 sin1 sin 1 sinθ θ

θ θ− + + + −

is equal to (a) 2secθ (b) 2secθ− (c) 2cosecθ (d) 2cosecθ− 36. If 60 ,θ = ° then 21 tan2 tan θ

θ+ is equal to

(a) 32 (b) 23 (c) 13 (d) 3 37. If sec and tan ,m nθ θ= = then ( )

( )1 1

m nm m n

+ +

+ is equal to

(a) 2 (b) mn (c) 2m (d) 2n 38. The expression 1 sin cos1 sin cosθ θ

θ θ+ −+ +

is equal to where 0, 2Aπ ∈

(a) 1 sin cos2 2θ θ− (b) 1 sin cos2 2θ θ

+ (c) tan 2θ (d) 22(cot 1)θ − . 39. If 1̀30A = ° and sin cos ,x A A= + then



(a) 0x < (b) 0x = (c) 0x > (d) 0x ≥ 40. If 3,2 2π π

α π π β< < < < and 3 3cos , sin2 5α β= − = − then 23 tan 2 tancot cosα βα β

++

is equal to (a) 320 (b) 522 (c) 425 (d) 727 41. If θ is an acute angle and 1tan ,7θ = then the value of 2 2

2 2cosec seccosec secθ θθ θ−+

is (a) 12 (b) 34 (c) 54 (d) 2 42. If A lies in the second quadrant and 3 tan 4 0,A+ = then the value of 2cot 5cos sinA A A− + is equal to (a) 5310− (b) 710− (c) 1710 (d) 2310 43. If 1tan and tan ,1 2 1x

x xα β= =

+ + then α β+ is where α and β are acute angles

(a) 0 (b) 4π (c) 3π (d) 2π 44. The value of cot 54 tan 20tan 36 cot 70° °

+° °

is (a) 0 (b) 1 (c) 2 (d) 3 45. If 5 1tan and tan , where 0 , ,6 11 2πα β α β = = < <

then

(a) 6πα β+ = (b) 4πα β+ = (c) 3πα β+ = (d) 2πα β+ = 46. The value of tan 70 tan 20tan 50° − °

° is equal to

(a) 0 (b) 1 (c) 2 (d) 3 47. The value of ( )tan 945− ° is (a) 3− (b) 3− (c) 13− (d) 1− 48. sin 200 cos 200° + ° is (a) negative (b) zero (c) positive (d) either zero or positive 49. If , , ,A B C D are the angles of a cyclic quadrilateral, then cos cos cos cosA B C D+ + + is equal to (a) 1− (b) 0 (c) 1 (d) 4 50. Evaluate tan 255 tan 345tan195 tan105°+ °

° − ° if cot15 m° =

(a) 11m

m

−+

(b) 22 1mm + (c) 2 11m

m

−+

(d) None of these 51. If 5cosec cot 2θ θ+ = , then the value of tanθ is (a) 1425 (b) 2021 (c) 2120 (d) 1516 52. ( ) ( )cos cos 90 sin sin 90θ θ θ θ°− − °− equals



(a) 1− (b) 0 (c) 1 (d) 2 53. If 1cos 2θ = − and 0 360θ° < < ° , then the values of θ are (a) 60° and 120° (b) 60° and 240° (c) 120° and 240° (d) 120° and 300° 54. 2sin 1 cos sin1 1 cos sin 1 cosθ θ θ

θ θ θ+

− + =+ −

equals (a) 0 (b) 1 (c) sinθ (d) cosθ 55. The value of the expression 6 6 2 2sin cos 3sin cosθ θ θ θ+ + equals (a) 0 (b) 1 (c) 2 (d) greater than 3 56. If cosec cot qθ θ− = , then the value of cotθ is (a) 221 q

q+ (b) 221 q

q− (c) 21 2qq− (d) 21 2qq+

57. If sec 2A = and 3 22 Aπ

π< < , then 1 tan cosec1 cot cosecA A

A A

+ ++ −

is equal (a) 2− (b) 1− (c) 0 (d) 12 58. If θ lies in the first quadrant and 5 tan 4θ = , then 5sin 3cossin 2cosθ θ

θ θ−+

is equal to (a) 0 (c) 114 (c) 314 (d) 514 59. The cotangent of the angles ,3 4π π and 6π are in (a) A.P. (b) G.P. (c) H.P. (d) None of these 60. 3cos cos3sin sin 3θ θ

θ θ+ 3−

is equal to (a) 2cotθ (b) 2cot 1θ + (c) 3cot θ (d) 4cot θ 61. The value of 3 3 3log tan1 log tan 2 .... log tan89° + ° + + ° is (a) 0 (b) 1 (c) 2 (d) 3 62. Which of the following is correct? (a) sin1 sin1° = (b) sin1 sin1180π° = (c) sin1 sin1° < (d) sin1 sin1° > 63. The value of tan1 tan 2 tan 3 .... tan 89° ⋅ ° ° ° is equal to (a) 1− (b) 1 (c) 2π (d) 2 64. If 12sin 013 2πθ θ = < <

and 3 3cos ,5 2πφ π φ

− = < <

, then ( )sin θ φ+ will be (a) 156− (b) 5665− (c) 5661− (d) 165 65. The value of 2

21 tan 151 tan 15− °+ °

is (a) 1 (b) 32 (c) 3 (d) 2



66. If 1 2tan cotkθ θ= , then ( )( )

1 21 2

coscos θ θθ θ

−

+ is equal to

(a) 11k

k

−+

(c) 11 k

k

−+

(c) 11k

k

+−

(d) 11 k

k

+−

67. The value of cos 480 sin150 sin 600 cos390° ⋅ ° + °⋅ ° is (a) 1− (b) 12− (c) 0 (d) 12 68. Find the value of k for which ( )2cos sin sin cos 1 0x x k x x+ + − = is an identity. (a) 2− (b) 1− (c) 1 (d) 2 69. The value of cot105° is (a) 3 2− (b) 2 3− (c) 2 3+ (d) 3 2+ 70. The expression ( ) ( ) ( )sin sin sinsin sin sin sin sin sinα β β γ γ α

α β β γ γ α− − −

+ + is equal to (a) 0 (b) 1 (c) ( )sin α β γ+ + (d) sin sin sinα β γ+ + 71. If 2sin sin 1x x+ = , then the value of 2 4cos cosx x+ is (a) 1 (b) 0 (c) 1− (d) 2 72. If ,cos sina b

x h y kθ θ

= + = + , then ( ) ( )

2 22 2a b

x h y k+

− − is equal to

(a) ab (b) aba b+

(d) 2ab (d) 1 73. The expression tan cottan cotθ ϕ

ϕ θ−−

is equal to (a) tantanθϕ (b) 1 tan1 tanθϕ+

+ (c) 1 tan1 tanθϕ−

− (d) tantanϕθ

74. The value of tan 5 tan15 tan 25 tan 45 tan 35 tan 55 tan 65 tan 75 tan85° ⋅ ° ⋅ ° ⋅ ° ⋅ ° ⋅ ° ⋅ ° ⋅ ° ⋅ ° is (a) ∞ (b) 0 (c) 1 (d) 1− 75. The value of ( ) ( ) ( ) ( )cos 54 cos 54 cos 36 cos 36θ θ θ θ°+ ° − + °− °+ is (a) cosθ (b) cos 2θ (c) cos3θ (d) sin 2θ 76. If 3 12sin , cos , where 5 13 2A B A

ππ= = < < and 32B

ππ < < , then the value of ( )tan A B+ is

(a) 1663 (b) 6316 (c) 6316− (d) 1663− 77. If (1 cos ) / (1 cos )x θ θ= − + then 22(1 )xx− is equal to: where 0, 2πθ ∈

where θ lies in first quadratnt (a) sinθ (b) cosθ (c) tanθ (d) cotθ .



78. The expression sin sin21 cos cos2θ

θ

θθ

+

+ + is equal to

(a) tan 2θ (b) cot 2θ (c) sin 2θ (d) cos 2θ 79. The value of 1 cos2 sin 21 cos2 sin 2θ θ

θ θ+ +− +

is equal to (a) cotθ (b) tanθ (c) secθ− (d) cosecθ 80. The value of 2 2 2cos4θ+ + is where , 4πθ θ ∈

(a) cosθ (b) 2cosθ (c) cos 2θ (d) 2cos 2θ 81. The value of sin 2 sin5 sincos 2 cos5 cosx x x

x x x

+ −+ +

is (a) cot 2x (b) tan x (c) tan 2x (d) cot x 82. The value of ( ) ( )

( ) ( )sin 1 sin 1cos 1 2cos cos 1n n

n n n

α αα α α+ − −

+ + + − is

(a) tanα (b) tan 2α (c) cot 2α (d) sec 2α 83. The value of sin 70 cos 40cos 70 sin 40°− °

° + ° is

(a) 3 (b) 13 (c) 13 (d) tan10° 84. The value of cos cos sin sinsin sin cos cos

n nA B A B

A B A B

+ + + − − , when n is odd, is

(a) 0 (b) 1− (c) 1 (d) 2cot 2n A B−

85. If cot cot 2α β = then ( )

( )coscos α β

α β+

− is

(a) 3 (b) 2 (c) 13− (d) 12 86. The value of cos 0 cos1 cos 2 ... cos179°+ ° + ° + + ° is (a) 0 (b) 1 (c) 1− (d) 1/ 2 87. If 90A B C+ + = ° then 2 2 2cos cos cosA B C+ + is (a) ( )2 1 cos cos cosA B C+ (b) 2 2sin sin sinA B C+ (c) ( )2 1 sin sin sinA B C− (d) ( )2 1 cos cos cosA B C− 88. If ( )tan A B p+ = , ( )tan A B q− = , then the value of tan 2A in terms of p and q is



(a) 1p q

pq

−+

(b) p q

p q

+−

(c) 1p q

pq

+−

(d) 11 pq

pq

−+

89. cos10 sin10cos10 sin10° + °

° − ° is equal to

(a) cot 35− ° (b) tan 35− ° (c) tan 55° (d) cot 55° 90. The value of 312sin 40 16sin 40° − ° is (a) 3 2− (b) 2 3− (c) 2 3 (d) 3 2 91. In a triangle ABC , let 1tan 2A = and 1tan 3B = . The value of angle C is ……. (a) 4π (b) 3π (c) 34π (d) 23π 92. If A lies in the third quadrant and 3 tan 4 0A− = , then 5sin 2 3sin 4cosA A A+ + is equal to (a) 245− (b) 0 (c) 245 (d) 485 93. If 8cos 17θ = and θ lies in the Ist quadrant, then the value of ( ) ( )cos 30 cos 45θ θ°+ + °−

( )cos 120 θ+ °− is (a) 23 3 1 117 2 2 −

−

(b) 23 3 1 117 2 2 +−

(c) 23 3 1 117 2 2 −

+

(d) 23 3 1 117 2 2 ++

94. 2 2sin 17.5 sin 72.5°+ ° is equal to (a) 2cos 30° (b) 2sin 45° (c) 2tan 45° (d) 2cos 90° 95. The value of 3 34sin cos 4cos sinA A A A− is equal to (a) sin 2A (b) sin 3A (c) cos3A (d) sin 4A 96. 3 3cos cos3 sin sin 3cos sinθ θ θ θ

θ θ− +

+ is equal to (a) 0 (b) 1 (c) 3 (d) 5 97. If 2 2tan 2 tan 1α β= + , then cos 2α is equal to (a) tan β− (b) 2tan β (c) sin β− (d) 2sin β− 98. If 2 ,A C B+ = then cos cossin sinC A

A C

−−

is equal to: (a) tan B (b) cot B (c) tan 2B (d) cot 2B . 99. If 1cos 7P = and 13cos 14Q = , where P and Q both are acute angles. Then, the value of P Q− is (a) 30° (b) 45° (c) 60° (d) 75° 100. sin120 cos150 cos 240 sin 330° ° − ° ° is equal to (a) 3 14+− (b) 1− (c) 1 (d) 23 101. 4 4cos sin24 24π π −

is equal to

(a) 5 12 2− (b) 5 14− (c) 3 12 2+ (d) 2 22+



102. The value of sin 300 tan 330 sec 420tan135 sin 210 sec315° ° °° ° °

is equal to (a) 13 (b) 12 (c) 2 (d) 3 103. 3 5 71 cos 1 cos 1 cos 1 cos8 8 8 8π π π π + + + +

is equal to: (a) 18 (b) 12 (c) 1 22 2+ (d) cos 8π . 104. Find the maximum value of sin sinx y⋅ if 90x y+ = ° . (a) 13 (b) 12 (c) 23 (d) 1 105. If tan tan mα β− = and cot cot nα β− = , then ( )cot α β− is equal to (a) 2 21 1

m n− (b) 2 21

m n− (c) 1 1

m n− (d) m n

n m−

106. If 1sec tan 1m

mθ θ

+− =

−, then cosθ is equal to

(a) 22 1mm − (b) 22 1mm +

(c) 22 11m

m

−+

(d) 22 11m

m

+−

107. If 1tan 3α = and 1tan 7β = then ( )2α β+ is equal to (where ,α β are acute) (a) 6π (b) 4π (c) 3π (d) 2π 108. If 212cot 31cosec 32 0θ θ− + = , then the value of sinθ is (a) 23± (b) 35 or 1 (c) 12± (d) 45 or 34 109. 1 1sec tan cosθ θ θ

−−

is equal to (a) 1 1sin sec tanθ θ θ

++

(b) 1 1sin sec tanθ θ θ−

+

(c) 1 1cos sec tanθ θ θ+

+ (d) 1 1cos sec tanθ θ θ

−+

110. If α is an acute angle and 22 sin 12x x

α⋅ + = , then tanα is

(a) 2 1x − (b) 2 1x + (c) 2 2x − (d) 2 12x − 111. sin sin 21 cos cos 2θ θ

θ θ+

+ + is equal to

(a) sinθ (b) cosθ (c) tanθ (d) cotθ 112. If 34 x

ππ< < then 212cot sinx

x+ is equal to

(a) sin cossinx x

x

− − (b) sin cossinx x

x

− + (c) sin cossinx x

x

− (d) sin cossinx x

x

+



113. cos1 cos 2 cos3 ..... cos180°+ ° + ° + + ° is equal to (a) 1− (b) 0 (c) 1 (d) 2 114. If angle C of a triangle ABC satisfies the equation 5cos 3 0C + = , then tanC and sinC are the roots of the equation (a) 215 8 16 0x x− − = (b) 215 8 16 0x x− + = (c) 215 8 16 0x x+ − = (d) 215 8 16 0x x+ + = 115. If ( ) ( ) 1sin 1, sin 2α β α β+ = − = then ( ) ( )tan 2 tan 2α β α β+ ⋅ + is equal to (where α and β are acute) (a) 1− (b) 0 (c) 1 (d) None of these 116. Which of the following is incorrect? (a) ( )2 2 2cot tan sec cosecθ θ θ θ+ = + (b) ( )2 2 2cot tan sec cosecθ θ θ θ+ = ⋅ (c) ( )2 2cot tan 4cot 2θ θ θ− = (d) None of these 117. The set of all values of x in the interval 0, 2π

for which 22sin 3sin 1 0x x− + ≥ is

(a) 0, 6π

(b) ,6 3π π

(c) 0, 4π

(d) ,4 3π π

118. If 2 2 2

2 2 21 sin sin sincos 1 cos cos4sin 4 4sin 4 1 4sin 4θ θ θ

θ θ θ θθ θ θ

+

+ =

+

, then sin 4θ is equal to (a) 1− (b) 12− (c) 12 (d) 1 119. 2 2sin sin8 2 8 2A Aπ π + − −

is equal to

(a) 3 sin2 A (b) 1 sin2 A (c) 1 sin2 A (d) 1 sin2 2 A 120. If tan b

xa

= , then the value of cos 2 sin 2a x b x+ is (a) a b− (b) a (c) b (d) a b+ 121. If tan cot mα α+ = , then 3 3tan cotα α+ is equal to (a) ( )2 3 3m m m− + (b) ( )2 3m m − (c) ( )2 3m m + (d) ( )2 3m m m− + 122. 2 2cos sin6 6π π

θ θ + − −

is equal to (a) 21 cos2 θ− (b) 0 (c) 1 cos 22 θ (d) 21 cos2 2θ 123. If ,A B C π+ + = then sin 2 sin 2 sin 2cos cos cos 1A B C

A B C

+ ++ + −

is equal to (a) 2cos cos cos2 2 2A B B C C A+ + +

(b) 4sin sin sin2 2 2A B C (c) 8cos cos cos2 2 2A B C (d) 4 tan tan tan2 2 2A B B C C A+ + +

124. If ,A B C π+ + = then sin 2 sin 2 sin 2 ,A B C+ + is equal to



(a) 2sin sin sinA B C (b) 2cos cos cosA B C (c) 4sin sin sinA B C (d) 4cos cos cosA B C 125. The expression 2cos cos3 cos5θ θ θ− − =…… (a) 2 38cos sinθ θ (b) 2cos 2 cos3θ θ (c) 2 316sin cosθ θ (d) 4sin 3 sin 2θ θ 126. If 2cos 2cos3 cos and 2sin 2sin 3 sin ,x x y x x y+ = + = then the value of cos 2x is (a) 78− (b) 18− (c) 18 (d) 78 127. If θ is an acute angle and 1sin 2 2x xθ −

= then tanθ is equal to (a) 2 1x − (b) 2 1x − (c) 2 1x + (d) 2 1x + 128. sin 3 sin 5 sin 7 sin 9 ....cos3 cos5 cos7 cos9θ θ θ θ

θ θ θ θ+ + +

=+ + +

(a) cot 5 tan 7θ θ (b) 2 tan 2θ (c) 2cot 6θ (d) tan 6θ 129. If ( )

cos sin ,sin cosAα α

αα α

= −

then ( ) ( ).A Aα β is equal to (a) ( )A α β− (b) ( ) ( )A Aα β− (c) ( ) ( )A Aα β+ (d) ( )A α β+ 130. The value of cos15 sin15°− ° is (a) 12− (b) 0 (c) 12 (d) 12 131. For all values of ,x the expression sin cosx x+ is always (a) 2≥ (b) 1≥ (c) 2≤ (d) 12≥ 132. ( ) ( )

( )

323tan / 3 tan / 31 3tan / 3A A

A

−

− is equal to

(a) sin A (b) 2 tanA A− (c) cot A (d) tan A 133. If 2cos ,C θ= then the value of the determinant 1 01 16 1

C

C

C

∆ = is (a) ( )24cos 2cos 1θ θ − (b) sin 4sin θ

θ (c) 22sin 2sin θ

θ (d) None of these

134. The expression cos 20 cos 40 , cos 60 cos80° ° ° ° is equal to (a) 14 (b) 38 (c) 512 (d) 116 135. If , ,α β γ are in A.P. then sin sincos cosα γ

γ α−−

is equal to (a) sinα (b) cosα (c) tan β (d) cot β 136. cos52 cos68 cos172° + ° + ° is equal to (a) 0 (b) 1 (b) 2 (d) None of these 137. If 2 ,A B C π+ + = then 2 2 2cos cos sinA B C+ − is equal to (a) 2cos cos sinA B C (b) 2sin sin cosA B C (c) 2cos cos cosA B C (d) 2sin sin sinA B C



138. If ( ) ( )2 4sec tan , sec tan ,x a y bθ θ θ θ= + = − then 4 2x y is equal to (a) 2 4a b (b) 4 2a b (c) 2 2 2tana b θ (d) secab θ 139. The value of 2 2(1 tan tan ) (tan tan )A B A B+ + − is: (a) 2 2cos cosA B (b) 2 2tan tanA B (c) 2 2cot cotA B (d) 2 2sec secA B . 140. The maximum value of the expression 3sin cos 2θ θ+ is (a) 137 (b) 1115 (c) 178 (d) 45 141. The value of 2 22sin sin15 30π π

− is (a) 5 18− (b) 5 16− (c) 5 13+ (d) 5 14+ 142. ( )

3 21 cos 2 1 12k

kπ

=

−

∑ is equal to (a) 12− (b) 0 (c) 12 (d) 32 143. The minimum value of ( ) 4 4sin cos , 0 ,2f x x x x

π= + ≤ ≤ is

(a) 14− (b) 12 2 (c) 14 (d) 12 144. If ,A B C π+ + = then 2 2 2sin sin sinA B C+ + is equal to (a) 2 cos cos cosA B C− (b) 1 2sin sin sinA B C− (c) 2 sin sin sinA B C+ (d) 2 2cos cos cosA B C+ 145. If 3 15, ; sin2 2 17π π

α π π β α< < < < = and 12tan ,5β = then the value of ( )sin β α− is (a) 171221− (b) 21221− (c) 21221 (d) 171221 146. The expression cos 2 2cosθ θ+ has values lying in the interval (a) 3 , 32 −

(b) 3 1,2 2 − (c) 1 , 22 −

(d) [ ]2, 2− 147. ( ) ( )2 2 2cos cos 120 cos 120α α α+ − ° + + ° is equal to (a) 0 (b) 12 (c) 1 (d) 32 148. If 35 , 15 and 40A B C= ° = ° = ° then tan . tan tan .tan tan . tanA B B C C A+ + is equal to (a) 0 (b) 1 (c) 2 (d) 3 149. The value of ( ) ( )6 6 4 42 sin cos 3 sin cos 1θ θ θ θ+ − + + is equal to (a) 0 (b) 2 (c) 4 (d) 6 150. sin 5sin θ

θ is equal to

(a) 4 216cos 12cos 1θ θ− − (b) 4 216cos 12cos 1θ θ− + (c) 4 216cos 12cos 1θ θ+ − (d) 4 216cos 12cos 1θ θ+ + 151. If 21 cos cos ...... 2 2,α α+ + + = − then ( )0α α π< < is



(a) 8π (b) 6π (c) 4π (d) 34π 152. 2 21 1cos 7 cos 372 2 ° − °

is equal to

(a) 34 (b) 12 (c) 12 (d) 12 2 153. The expression 3 5 7 9tan tan tan tan tan20 20 20 20π π π π π

20 is equal to

(a) 23 (b) 1 (c) 2 3 (d) 2 154. The value of 2 2 2 5cos cos cos12 4 12π π π

+ + is (a) 3 32+ (b) 32 (c) 23 3+

(d) 23 155. If 0A B C+ + = , then sin sin sinA B C+ + is equal to (a) 4sin sin sin2 2 2A B C

− (b) 1 4cos cos cos2 2 2A B C−

(c) 2sin sin sin 12 2 2A B C− − (d) 12cos cos cos2 2 2 2A B C

− + 156. The value of cos35 cos85 cos155° + ° + ° is: (a) 12− (b) 0 (c) 12 (d) 1. 157. If ( ) ( ) 1sin 1, tan 3A B C A B+ + = − = and ( )sec 2,A C+ = then where , ,A B C are acute (a) 45 , 15 , 30A B C= ° = ° = ° (b) 60 , 30 , 0A B C= ° = ° = ° (c) 90 , 60 , 30A B C= ° = ° = ° (d) 30 , 60 , 0A B C= ° = ° = ° 158. Which of the following statements is incorrect? (a) sin sin sin 4cos cos cos when2 2 2A B C

A B C A B C π+ + = + + = (b) sin sin sin 4sin sin sin when 22 2 2A B C

A B C A B C π+ + = + + = (c) sin sin sin 4sin sin sin when 02 2 2A B C

A B C A B C+ + = + + = (d) sin 2 sin 2 sin 2 4sin sin sin whenA B C A B C A B C π+ + = + + = 159. The expression sin sin sin3 3π π

θ θ θ − +

is equal to: (a) 1 sin 23 θ (b) 21 sin4 θ (c) 1 sin 33 θ (d) 1 sin 34 θ . 160. If θ and φ are angles in the 1st quadrant such that 1tan 7θ = and 1sin ,10φ = then (a) 2 30θ φ+ = ° (b) 2 45θ φ+ = ° (c) 90θ φ+ = ° (d) 2 90θ φ+ = ° 161. In ,ABC∆ the value of ( )cosec sin cos cos sinA B C B C+ is equal to



(a) 1− (b) 0 (c) 1 (d) 12 162. The expression cos 6 cos 4 cos 2 1θ θ θ+ + + is equal to (a) 2sin sin 3θ θ (b) 24sin 3θ (c) 4cos cos 2 cos3θ θ θ (d) 2sin 3 cos3θ θ 163. If ( ) ( )2 2 2 2sin cosec cos sec tan cot ,kα α α α α α+ + + = + + then k is equal to (a) 3 (b) 5 (c) 7 (d) 9 164. The value of sin 70 cos 40cos 70 sin 40°+ °

° + ° is equal to

(a) 12 (b) 13 (c) 1 (d) 3 165. The value of 2 2cos sin4 4π π

θ θ + − −

is (a) 0 (b) cosθ (c) sin 2θ (d) cos 2θ 166. The ratio of the maximum value of 2 7sin cos 4θ θ− + to its minimum value is (a) 2 (b) 4 (c) 7 (d) 114 167. The maximum value of 5sin 3sin 33πθ θ + + +

is

(a) 9 (b) 10 (c) 11 (d) 12 168. The expression tan 20 .tan 40 . tan 60 .tan80° ° ° ° is equal to (a) 3 (b) 2 3 (c) 3 (d) 3 3 169. The expression cos cos cos3 3π π

θ θ θ − +

is equal to: (a) 1 cos 23 θ (b) 21 cos4 θ (c) 1 sin 33 θ (d) 1 cos34 θ . 170. sin163 cos347 sin 73 sin167° ° + ° ° is equal to (a) 0 (b) 12 (c) 1 (d) None of these 171. The maximum value of 2 sin 2cos4 4π π

θ θ + + + −

is: (a) 2 (b) 3 2+ (c) 2 5+ (d) 7 . 172. If 4A B

π− = , then ( )( )1 tan 1 tanA B+ − is equal to

(a) 2− (b) 1− (c) 1 (d) 2 173. The value of sin 600 cos330 cos120 sin150° °+ ° ° is (a) 1− (b) 12 (c) 1 (d) 32 174. If 2A B C

π+ + = , find sin 2 sin 2 sin 2A B C+ + .

(a) 4cos sin sinA B C (b) 4sin sin sinA B C (c) 4cos cos cosA B C (d) 4cos cos sinA B C 175. The value of sin12 .sin 48 .sin 54° ° ° is:



(a) 18 (b) 16 (c) 13 (d) 12 . 176. If 2 2cos cos cos3 3

x y z

π πθ θ θ= =

+ −

, then the value of x y z+ + is (a) 1 (b) 0 (c) 12 (d) 12− 177. 2 2sin sinsin cos sin cosA B

A A B B

−−

is equal to (a) ( )tan A B− (b) ( )cot A B− (c) ( )tan A B+ (d) ( )cot A B+ 178. The value of sin 85 sin 35cos 65°− °

° is

(a) 1− (b) 0 (c) 1 (d) 2 179. ( ) ( ) ( )

( ) ( ) ( )sin 660 tan 1050 sec 420cos 225 cosec 315 cos 510− ° ° − °

° ° ° is equal to

(a) 43 (b) 23 (c) 32 (d) 34 180. If 3cos 4A = , then the value of 532sin sin2 2A A is equal to (a) 11− (b) 11− (c) 11 (d) 11 181. The minimum value of ( )2 24 tan 9cotθ θ+ is (a) 4 (b) 6 (c) 9 (d) 12 182. Which is not true in a triangle? (a) sin sin cos cos sinA B C B C= + (b) cos sin sin cos cosC A B A B= − (c) tan tan tan tan tan tanA B C A B C+ + = ⋅ ⋅ (d) cot cot cot cot cot cotA B C A B C+ + = ⋅ ⋅ 183. If cos cos cos 0x y α+ + = and sin sin sin 0x y α+ + = , then cot 2x y+

is equal to (a) sinα (b) cosα (c) cotα (d) sin 2x y+

184.

( )2 24sec xy

x yθ =

+ is true if and only if

(a) 0, 0x y≠ ≠ (b) , 0x y x= ≠ (c) x y= (d) 0x y+ ≠ 185. If 180A B C+ + = ° , then tan tan2 2A B

∑ is equal to (a) 0 (b) 1 (c) 2 (d) 3 186. If tan tan tan tan 33 3 k

π πθ θ θ θ ⋅ + ⋅ − + =

, then the value of k is

(a) 1− (b) 13 (c) 3 (d) None of these 187. If sin sinA n B= , then 1 tan1 2n A B

n

− + +

is equal to



(a) sin 2A B−

(b) tan 2A B−

(c) cot 2A B−

(d) cos 2A B−

188. If 180A B C+ + = ° , then cos 2 cos 2 cos 2A B C+ + is equal to (a) 1 4sin sin sinA B C− + (b) 1 4cos cos cosA B C− (c) 1 4cos cos cosA B C+ (d) None of these 189. If 0, 2πα ∈

, then 22 2tan

x xx x

α+ +

+ is always greater than or equal to

(a) 1 (b) 2 (c) 2 tanα (d) 2sec α 190. The quadratic equation whose roots are 2sin 18° and 2cos 36° is (a) 216 12 1 0x x− − = (b) 216 12 1 0x x− + = (c) 216 12 1 0x x+ − = (d) 216 12 1 0x x+ + = 191. If 180A B C+ + = ° , then tan tan2 2A B

⋅∑ is equal to (a) 0 (b) 1 (c) 2 (d) 3 192. If sin sinα β= and cos cosα β= , then: (a) sin 02α β− =

(b) cos 02α β− =

(c) sin 02α β+ =

(d) cos 02α β+ =

.

193. If 270 ,A B C+ + = ° then cos 2 cos 2 cos 2A B C+ + 4sin sin sinA B C+ is equal to: (a) 0 (b) 1 (c) 2 (d) 3. 194. sec8 1sec 4 1A

A

−−

is equal to (a) cot 8cot 2AA (b) tan 8tan 2AA (c) tan 2tan 8AA (d) None of these 195. The expression sin 36 sin 72 sin108 sin144° ° ° ° is equal to (a) 38 (b) 516 (c) 712 (d) 34 196. The maximum value of 2 2 2cos cos3 3x x

π π + + +

is (a) 32− (b) 23− (c) 23 (d) 32 197. If secx h a θ= + and cosecy k b θ= + , then (a)

( ) ( )

2 22 2 1a b

x h y k+ =

− − (b) ( ) ( )2 2

2 2 1x y y k

a b

− −+ =

(c) ( ) ( )

2 22 2 1a b

x h y k− =

− − (d) ( ) ( )2 2

2 2 1x y y k

a b

− −− =

198. The extreme values of ( )2 2 24cos cos cos3 3x x xπ π + −

are

(a) 1, 1− (b) 2, 2− (c) 3, 3− (d) 4, 4− 199. The expression ( )4 46 sin 5 sin 2ππ θ θ

+ + +

( )6 634 sin sin 32π θ π θ − + + +

is equal to

(a) 2 (b) 2sin 2θ (c) 2 21 sin cos2 θ (d) 1 cos 22 θ



200. If tan cos tanβ θ α= , then 2tan 2θ

is equal to (a) ( )

( )coscos α β

α β+

− (b) ( )2cot β α− (c) ( )

( )sinsin α β

α β−

+ (d) ( ) ( )tan cotα β α β− +

201. The value of 2 23 4cos cos5 5π π+ is equal to

(a) 34 (b) 45 (c) 54 (d) 52 202. If ( )

( )cos 1sin 1p

p

α βα β− +

=+ −

, then p is equal to (a) tan tan4 4π π

α β − −

(b) tan tan4 4π πα β − +

(c) tan tan4 4π πα β + −

(d) tan tan4 4π π

α β + +

203. If sin 4 cos 2 cos 4 sin 2 , 0 4A A A A A

π − = − < <

, then the value of tan 4A is (a) 1 (b) 13 (c) 3 (d) 3 13 1+− 204. In a 2 3, cos cos2 2B C A A B

ABC+ + − ∆ +

is equal to

(a) 1− (b) 0 (c) 1 (d) 2 205. Evaluate 2 2cot 36 cot 72° ° (a) 5 14− (b) 15 (c) 5 14+ (d) 52 206. If tan tanA B x− = and cot cot ,B A y− = then cot( )A B− is equal to: (a) 1 1

x y− (b) y x

y x

−+

(c) 1 1x y+ (d) 2y x

y x

−−

. 207. If 3 1tan ,3 1A

−=

+ (where A is acute angle) then, cos A =

(a) 3 12 2− (b) 3 12 2+ (c) 3 1+ (d) 3 1− 208. If 1tan ,7θ = (where θ is acute angle) then, 2 2

2 2cosec seccosec secθ θθ θ−+

(a) 12 (b) 34 (c) 56 (d) 78 209. If tan cot 2,θ θ+ = (where θ is acute angle) then, sin cosθ θ+ = (a) 5 (b) 2 (c) 3 (d) 2 210. If 2cos tan 2cos tan 1 0,A B A B⋅ − − + = (where A and B are acute) then the measures of angles A and B are respectively (a) 60 , 45° ° (b) 30 , 45° ° (c) 45 , 75° ° (d) 45 , 60° °



211. 3 3 3 3sin cos sin cossin cos sin cosθ θ θ θθ θ θ θ+ −

+ =+ −

(a) 0 (b) 1 (c) 2 (d) 3 212. ( ) ( )sec tan 1 sec tan 1θ θ θ θ+ − − + = (a) 2sinθ (b) 2cosθ (c) 2secθ (d) 2 tanθ 213. If ( ) ( )2 2sin cosec cos secθ θ θ θ+ + + 2 2tan cot kθ θ= + + then, k = (a) 1 (b) 3 (c) 5 (d) 7 214. If 4 4 2 2 2sin cos sin .cos 1 ,uθ θ θ θ+ + = − then u = (a) sin cosθ θ⋅ (b) sec cosecθ θ⋅ (c) sec tanθ θ⋅ (d) cosec cotθ θ⋅ 215. If ( )( )sin cos 1 sin cos sin cosn nθ θ θ θ θ θ+ − ⋅ = + , then n = (a) 0 (b) 1 (c) 2 (d) 3 216. If 2 2cos sin ,p q pθ θ⋅ = − ⋅ (where θ is acute angle) then sinq θ⋅ = (a) p (b) 2 2p q− (c) 2 2 cosp q θ+ ⋅ (d) None of these 217. If sin 3 cos ,θ θ= ⋅ (where θ is acute angle) then, ( )2 sin cos 1θ θ+ − = (a) 3 (b) 2 (c) 1 (d) 0 218. If sec cos 2,θ θ+ = then 2 4sec secθ θ− = (a) 1 (b) 2 4cos cosθ θ+ (c) 1− (d) 4 2cos cosθ θ− 219. If tan cot 2,θ θ+ = then 2 3tan tanθ θ− = (a) 3 2cot cotθ θ− (b) 1 (c) 2 3cot cotθ θ+ (d) 2 220. If 1 sin sin cos cos 0α β α β+ ⋅ − ⋅ = then tan .cotα β = (a) 3− (b) 2− (c) 1− (d) 0 221. If , ,A B C are the angles of a triangle, then tan tan cos cosec2 2 2 2A B C A B C+ +

⋅ + ⋅ = (a) 1 (b) 2 (c) 0 (d) 3 222. If cos cos , cos sin ,x r A B y r A B= ⋅ ⋅ = ⋅ ⋅ sin ,z r A= ⋅ then 2 2 2x y z+ + = (a) ( )2 2 2sin cosr A B+ (b) ( )2 2 2cos sinr A B+ (c) 2r (d) 22r 223. If 2 2cos sinx a bθ θ= ⋅ + ⋅ and ( )( ) 2 2 2sin cos ,x a b x c θ θ− − = ⋅ ⋅ then c = (a) a b− or b a− (b) a b+ (c) 2 2a b− or 2 2b a− (d) 2 2a b+ 224. If ( ) ( )cos cosx A B A B= + ⋅ − and ( ) ( )sin sin ,y A B A B= + ⋅ − then x y− = (a) sin 2A (b) cos 2A (c) sin 2B (d) cos 2B 225. If sin cos3 3x A B

π π = + ⋅ +

and cos sin ,3 3y A Bπ π = + ⋅ +

then x y− =

(a) ( )sin A B− (b) ( )cos A B− (c) ( )sin A B+ (d) ( )cos A B+ 226. 1 tan tan 2AA

+ ⋅ =

(a) sin A (b) cos A (c) cosec A (d) sec A 227. If ( )cos cos sin ,x A B A B⋅ ⋅ = − ( )cos cos sin ,y B C B C⋅ ⋅ = − ( )cos cos sin ,z A C C A⋅ = − then x y z+ + =



(a) 1− (b) 0 (c) ( )sin A B C+ + (d) ( )cos A B C− − 228. If ( )sin cos ,x B C A= − ⋅ ( )sin cos ,y C A B= − ⋅ and ( )sin cos ,z A B C= − ⋅ Then x y z+ + = (a) ( )sin A B C− − (b) ( )cos A B C− − (c) 0 (d) 1− 229. If 2sin cos ,3 6x x

π π + = −

then tan x = (a) 1/ 3 (b) 1/ 3− (c) 3 (d) 3− 230. ( ) ( )cos cos 120 cos 120x x x+ °− + °+ = (a) 1− (b) 0 (c) 3 (d) 1/ 3 231. ( ) ( )tan 45 tan 45A A° + ⋅ °− = (a) 1− (b) 0 (c) 1 (d) 2 232. tan 75 tan15° + ° = (a) 1 (b) 2 (c) 3 (d) 4 233. If tan , tan ,a x b y= = 21c a= + and 21d b= + (where ,x y ∈ 1st quadrant) then ( )sincd x y⋅ + = (a) a b− (b) a b+ (c) c d− (d) c d+ 234. 2 2cos cos4 4π π

θ θ − + + =

(a) sin 2θ (b) cos 2θ (c) 1 (d) 0 235. 3 9 11cos cos cos cos12 12 12 12π π π π

+ + + = (a) cos 2π (b) cosπ (c) 6cos 12π (d) 5cos 12π 236. 3 5 7 9cot cot cot cot cot20 20 20 20 20π π π π π

⋅ ⋅ ⋅ = (a) 1 (b) 2 (c) 3 (d) 4 237. 2 2 2 23 5 7sin sin sin sin8 8 8 8π π π π

+ + + = (a) 1 (b) 2 (c) 3 (d) 4 238. If 1 sin cos ,1 sin cos 2f

θ θ θθ θ

+ + = + − then 2f

θ =

…….. (a) sin 2θ

(b) cos 2θ

(c) tan 2θ

(d) cot 2θ

239. ( )

( )

221 tan 451 tan 45 A

A

− °−=

+ °−

(a) sin 2A (b) cos 2A (c) tan 2A (d) cot 2A 240. sin 3 cos3sin cosθ θ

θ θ− =

(a) sin 2θ (b) cos 4θ (c) 1 (d) 2 241. sin 2 cos 2sin cosθ θ

θ θ− =

(a) secθ (b) tanθ (c) 1 (d) 2



242. 3 34sin cos 4cos sinθ θ θ θ⋅ − ⋅ = (a) 4cosθ (b) cos 4θ (c) 4sinθ (d) sin 4θ 243. 2 41 8cos 8cosθ θ− + = (a) sin 4θ (b) cos 4θ (c) 4 4sin cosθ θ+ (d) 4 4sin cosθ θ− 244. If cos3 cos5 cos 7 cos15A A A A+ + + 4cos cos cosmA nA pA= ⋅ ⋅ then, the value of m n p+ + is (a) 14 (b) 15 (c) 16 (d) 17 245. If sin 40 cos 70 cos80 ,k° − ° = ⋅ ° then k = (a) 1 (b) 2 (c) 3 (d) 4 246. cos 20 cos100 cos140°+ ° + ° = (a) cos130° (b) cos 45° (c) 0 (d) None of these 247. cos55 cos65 cos175° + ° + ° = (a) 0 (b) 1 (c) sin18° (d) cos36° 248. sin10 sin 20 sin 40 sin 50°+ ° + ° + ° = (a) sin15 sin 75° + ° (b) cos15 cos75°+ ° (c) sin 70 sin80°+ ° (d) cos 70 cos80°+ ° 249. cos 21 sin 21cos 21 sin 21° − °

=° + °

(a) tan 21° (b) cot 66° (c) tan 42° (d) cot 42° 250. sin 7 sin sin11 sin 3θ θ θ θ⋅ + ⋅ = (a) sin 4 sin10θ θ⋅ (b) cos 4 cos10θ θ⋅ (c) sin 8 sin14θ θ⋅ (d) cos8 cos14θ θ⋅ 251. 2 3 4 516cos cos cos cos cos12 12 12 12 12π π π π π (a) 3 (b) 23 (c) 32 (d) 2 252. If ,A B C π+ + = then sin 2 sin 2 sin 2A B C+ − = (a) 4sin cos cosA B C⋅ ⋅ (b) 4cos sin cosA B C⋅ ⋅ (c) 4cos cos sinA B C⋅ ⋅ (d) 4cos cos cosA B C⋅ ⋅ 253. ( )sin 2 sin 2 sin 2A B A B+ + − = (a) ( )4sin .sin sinA B A B⋅ − (b) ( )4sin cos cosA B A B⋅ ⋅ − (c) ( )4cos sin cosA B A B⋅ ⋅ − (d) ( )4cos cos sinA B A B⋅ ⋅ − 254. 4 4tan cot 2A A+ + = (a) 2 21 sec cosecA A+ + (b) 2 22 sec cosecA A+ + (c) 2 2sec cosec 2A A⋅ − (d) 2 23 sec cosecA A− ⋅ 255. If tan tan .4 4 m

π πθ θ + − − =

( )tan ,nθ then

(a) m n< (b) m n= (c) m n> (d) 1nm < 256. 2 2sin cos4 2 4 2π θ π θ + − + =

(a) sinθ (b) cosθ (c) sin 2θ (d) cos 2θ 257. If cot ,a

bθ = then cos 2 sin 2a bθ θ⋅ + ⋅ =

(a) 2 2a b+ (b) 2 2a b− (c) a (d) b


21 Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1, Ph.: 0651-2562523, 9835508812, 8507613968

258. The expression 9 3 52cos cos cos cos13 13 13 13π π π π+ + is:

(a) – 1 (b) 0 (c) 1 (d) None of these. 259. The angle subtended at the centre of a circle of radius 3 m by an arc of length 1 m is equal to (a) 20° (b) 60° 13 rad (d) 3 rad 260. ( ) ( )cos 540 sin 630θ θ°− − °− = (a) 0 (b) 2cosθ (c) 2sinθ (d) sin cosθ θ+ 261. If tan 25 ,x = ° then tan155 tan1151 tan155 tan115° − °

=+ °⋅ °

(a) 21 2 xx− (b) 21 2 xx+ (c) 2

211 x

x

+−

(d) 2211 x

x

−+

262. If ( ) ( )sin 120 sin 120 ,α β°− = °− (where 0 ,α β π< < ) then (a) 3πα β+ = (b) α β= or 3πα β+ = (c) α β= (d) 0α β+ = 263. If 3sin 2 5 4cos 2 ,θ θ= + then tanθ = (a) 1 (b) 3 (c) 4 (d) 5 264. If cot 20 ,p = ° then tan160 tan1101 tan160 tan110° − °

=+ °⋅ °

(a) 2 12p p− (b) 2 12p p+ (c) 21 2 pp− (d) 221 p

p+

265. If ( )( )

sin ,sin x y a b

x y a b

+ +=

− − then tantan xy =

(a) ba

(b) ab

(c) ab (d) None of these 266. Which of the following number is rational ? (a) sin15° (b) cos15° (c) sin15 cos15° ⋅ ° (d) sin15 cos 75° ⋅ ° 267. If angle α is in third quadrant, then 1 cos 1 cos1 cos 1 cosα α

α α− +

+ =+ −

(a) 2cosecα (b) 2 cosecα− (c) cosecα (d) cosecα− 268. If ( )cot 0,α β+ = (where ,α β ∈ 1st quadrant), then ( )sin 2α β+ = (a) sinα− (b) sin β (c) cosα (d) cosβ 269. If cosec cot pθ θ− = , then cosecθ = (a) 1

pθ + (b) 1

pθ − (c) 1 12 p

p

+

(d) 1 12 p

p

−

270. If sec tan ,xθ θ− = (Where ( )0, 1x∈ and θ lies in 1st quadrant) then sinθ = (a) 2

211 x

x

+−

(b) 22 11x

x

−+

(c) 2 21xx

+ (d) 2211 x

x

−+

271. cos105 sin105° + ° = (a) sin 30° (b) cos 45° (c) sin 60° (d) cos90°



272. If 3cos ,4A = then 2 516cos 32sin sin2 2 2A A A − ⋅ =

(a) 4− (b) 3− (c) 3 (d) 4 273. If ( )sin 1α β+ = and ( ) 1sin ,2α β− = (where α and β are acute angles), then ( ) ( )tan 2 tan 2α β α β+ ⋅ + = (a) 1− (b) 0 (c) 1 (d) 2 274. If cos 25 sin 25 ,p° + ° = then cos50° = (a) 22 p− (b) 22 p− − (c) 22p p⋅ − (d) 22p p− ⋅ − 275. Which of the following is correct ? (a) csin1 sin1° > (b) csin1 sin1° < (c) csin1 sin1° = (d) ( ) csin1 /180 sin1π° = 276. If x is in the second quadrant, then 1 sin 1 sin1 sin 1 sinx x

x x

− ++ =

+ −

(a) 2sec x (b) 2sec x− (c) 2cosec x (d) 2cosec x− 277. sin cos3 3A B

π π + ⋅ +

cos sin ...3 3A Bπ π − + ⋅ + =

(a) ( )cos A B− (b) ( )sin A B− (c) ( )cos A B+ (d) None of these 278. If 0 2A

π< < and 0 2B

π< < , then angle ( )A B− lies surely in ….. quadrant

(a) First (b) Second (c) Third (d) None of these 279. If sin 3 / 5A = and cos 9 / 41,B = (where ,A B are both lie in the first quadrant), then ( )sin ...A B− = (a) 133 / 205− (b) 84 / 205− (c) 124 / 205 (d) None of these \ 280. ( )1 tan tan / 2 ....A A+ ⋅ = (a) sec A (b) tan A (c) cosec A (d) cot A 281. For any angle , , ,A B C ( ) ( ) ( )sin sin sincos cos cos cos cos cosA B B C C A

A B B C C A

− − −+ + =

⋅ ⋅ ⋅

(a) 0 (b) ( )sin A B C− − (c) ( )tan A B C− − (d) None of these 282. If ( ) ( )2sin 60 cos 30x x+ ° = − ° , then tan x =……. (a) 3 (c) 2 (c) 2− (d) 3− 283. If ( ) ( )sin 2 sin ,θ φ θ φ+ = ⋅ − then tan tan ,Kθ φ= ⋅ where ...K = (a) 1 (b) 2 (c) 3 (d) 4 284. If 3 tan tan 1,θ φ⋅ = then ( )

( )cos ...cos θ φ

θ φ−

=+

(a) 1 (b) 2 (c) 3 (d) 4 285. cot 54 tan 20 ...tan 36 cot 70° °

+ =° °

(a) 1 (b) 2 (c) 0 (d) 3 286. 2 2cos cos ...4 4π π

θ θ − + + =



(a) 1 (b) 2 (c) 3 (d) None of these 287. 3 5 7 9cot cot cot cot cot ...20 20 20 20 20π π π π π

⋅ ⋅ ⋅ ⋅ = (a) 1− (b) 0 (c) 1 (d) None of these 288. 2 2 2 2sin 5 sin 10 sin 15 ..... sin 90 ...°+ °+ °+ + ° = (a) 16 2 (b) 17 2 (c) 18 2 (d) 19 2 289. If sec 13 / 5,θ = − where 90 180 ,θ° < < ° then sin 2 ...θ = (a) 102 /191 (b) 120 /199 (c) 120 /169− (d) 119 /169− 290. If cos 3 / 7,θ = and θ lies in the fourth quadrant, then ( )cos / 2 ...θ = (a) ( )2 / 7 (b) ( )2 / 5− (c) ( )5 / 7− (d) None of these 291. For any angle ,A 4sin sin sin ...3 3A A A

π π ⋅ − ⋅ + =

(a) sin 2A (b) sin 3A (c) sin 4A (d) None of these 292. cos 20 cos 40 cos60 cos80° ⋅ ° ⋅ ° ⋅ ° =…… (a) 1/ 2 (b) 21/ 2 (c) 31/ 2 (d) 41/ 2 293. tan 20 tan 40 tan 60 tan 80 ...° ⋅ ° ⋅ ° ⋅ ° = (a) 1 (b) 2 (c) 3 (d) 4 294. If tan 3α = , then 2sin 2 3cos 24sin 2 5cos 2α α

α α−

=+

….. (a) 1/ 3 (b) 4 / 9 (c) 9 / 4− (d) 4 / 9− 295. 2 2 2 23 5 7cos cos cos cos ...8 8 8 8π π π π

+ + + = (a) 1 (b) 2 (c) 3 (d) 4 296. cos 40 cos80 cos160 cos 240 ...° + ° + °+ ° = (a) 2− (b) 1/ 2− (c) 3− (d) 1/ 3− 297. sin 70 cos 40 ...cos 70 sin 40°+ °

=°+ °

(a) 2 (b) 1/ 2 (c) 3 (d) 1/ 3 298. If 3sin 5sinα β= , then tan tan ...2 2α β α β+ − ÷ =

(a) 1 (b) 2 (c) 3 (d) 4 299. cos52 cos68 cos172 ...° + ° + ° = (a) 1− (b) 0 (c) 1 (d) None of these 300. If α is any real number, then 4 2 2 2

4 2 2 2sin sin cos cos ..cos sin cos sinα α α αα α α α+ ⋅ +

=+ ⋅ +

(a) 1 (b) 2 (c) 3 (d) None of these 301. If α is any real number, then ( ) ( ) ( )2 2 2 2sin cosec cos sec tan cot ...α α α α α α+ + + − + = (a) 3 (b) 5− (c) 7 (d) 2 22sin 3cosα α+ 302. sin163 cos347 sin 73 sin167 ....° ⋅ ° + °⋅ ° = (a) 2 (b) 1/ 2 (c) 3 (d) 1/ 3



303. If 99 99 99sin sin sin 3,a b c+ + = (where , ,a b c are acute angles) then 100 100 100cos cos cos ...a b c+ + = (a) 3− (b) 2− (c) 1− (d) 0 304. 2 2sin sinsin cos sin cosA B

A A B B

−=

⋅ − ⋅……

(a) ( )sin A B− (b) ( )cos A B+ (c) ( )tan A B+ None of these 305. If ( )sin 1,A B C+ + = ( )tan 1/ 3A B− = and ( )sec 2,A C+ = (where, , ,A B C∈ 0, 2π

) then

(a) 90 , 60 , 30A B C= ° = ° = ° (b) 120 , 60 , 0A B C= ° = ° = ° (c) 60 , 30 , 0A B C= ° = ° = ° (d) None of these 306. If ABC∆ is a right-angled at A then 2 2cos cosB C+ = (a) 2− (b) 1− (c) 1 (d) 0 307. For 0 2πθ≤ ≤ and 2 2

0 0cos , sinn n

n n

x yθ θ∞ ∞

= =

= =∑ ∑ , and 2 20 cos sinn n

n

z θ θ∞

=

= ⋅∑ (a) xyz xz y= + (b) xyz yz x= + (c) xyz xy z= + (d) x y z xyz+ + = 308. The circular wire of diameter 10 cm is cut and placed along the circumference of a circle of diameter 1 m. The angle subtended by the wire at the centre of the circle is equal to (a) 4π rad (b) 3π rad (c) 5π rad (d) 10π rad 309. The maximum value of 3cos 4sinθ θ+ is (a) 3 (b) 4 (c) 5 (d) None of these 310. If sin cos mθ θ+ = and sec cosec ,nθ θ+ = then ( ) ( )1 1n m m+ − is equal to (a) m (b) n (c) 2m (d) 2n 311. What is the value of ( ) ( )sin 420 cos390 cos 300 sin 330 ?° ⋅ ° + − ° ⋅ − ° (a) 0 (b) 1 (c) 2 (d) 1− 312. The value of ( ) ( )cot 45 cot 45θ θ° + ° − is (a) 1− (b) 0 (c) 2 (d) 1− 313. The value of ( ) ( )sin sin 60 sin 60A A A° − ° + is equal to (a) sin 3A (b) sin 32 A (c) sin 34 A (d) None of these 314. What is the value of sin cos tan cos sin cotA A A A A A+ ? (a) sin A (b) cos A (c) tan A (d) 1 315. What is the maximum value of 3cos 4sin 5x x+ + ? (a) 5 (b) 7 (c) 10 (d) 12 316. What is the value of sin 1 cos1 cos sinx x

x x

++

+ ?

(a) 2 tan x (b) 2 cosec x (c) 2cos x (d) 2sin x 317. What is the maximum value of sin 3 cos 2 cos3 sin 2θ θ θ θ+ ? (a) 1 (b) 2 (c) 4 (d) 10 318. The value of cos12 sin12 sin147cos12 sin12 cos147° − ° °

+° + ° °

is equal to (a) 1 (b) 1− (c) 0 (d) None of these



319. One of the angles of a triangle is 12 rad and the other is 99° The third angle in radian measure is …. (a) 9 10π

π− (b) 90 1007π π− (c) 90 10π

π− (d) None of these

320. If 2 2sec cos ,y θ θ= + where 0 ,2πθ< < then which one of the following is correct ? (a) 0y = (b) 0 2y≤ ≤ (c) 2y ≥ (d) None of these 321. If 3tan 4A = and 12tan ,5B = − then how many values can ( )cot A B− have depending on the actual values of A and B ? (a) 1 (b) 2 (c) 3 (d) 4 322. What is the value of sin cos tan ,sec cosec cotθ θ θ

θ θ θ+ −+ −

when 34πθ = ? (a) 0 (b) 1 (c) 1− (d) None of these 323. Which one of the following is correct ? (a) sin1 sin1° > (b) sin1 sin1° < (c) sin1 sin1° = (d) sin1 sin1180π° = 324. What is the value of cos15 sin15 cos 45 cos15cos 45 sin 45 sin 45 sin15° ° ° °

×° ° ° °

(a) 14 (b) 22 (c) 14− (d) 34− 325. The angle A lies in the third quadrant and it satisfies the equation ( )24 sin cos 1x x+ = What is the measure of the ?A∠ (a) 225° (b) 240° (c) 210° (d) None of these 326. If 1 1cos ,2 x

xθ = +

then 2 21 12 x

x

+

, is equal to (a) sin 2θ (b) cos 2θ (c() tan 2θ (d) scθ 327. The value of x for the maximum values of 3 cos sin ,x x+ is (a) 30° (b) 45° (c) 60° (d) 90° 328. If , , ,A B C D are the angles of a cyclic quadrilateral, then cos cos cos cosA B C D+ + + is equal to (a) ( )2 cos cosA C+ (b) ( )2 cos cosA B+ (c) ( )2 cos cosA D+ (d) 0 329. If 41 ,12A

π= then what is the value of 2

21 3tan ?3tan tanAA A

−−

(a) 1− (b) 1 (c) 13 (d) 3 330. The length of arc of a circle of radius 5 cm subtending a central angle measuring 15° is ? (a) 512π cm (b) 712π cm (c) 12π cm (d) 5π cm 331. What is the value of 1 sin10 sin 50 sin 70− ° ° ° ? (a) 18 (b) 38 (c) 58 (d) 78



332. The sines of two angles of a acute angled triangle are equal to 513 and 99 .101 What is the cosine of the third angle ? (a) 2551313 (b) 2651313 (c) 2751313 (d) 7701313 333. sin cosx θ θ= and sin cosy θ θ= + are satisfied by which one of the following equations ? (a) 2 2 1y x− = (b) 2 2 1y x+ = (c) 2 2 1y x− = − (d) 2 2 1y x+ = − 334. IF 4 4sin cos ,x x p− = then which one of the following is correct? (a) 1p = (b) 0p = (c) 1p > (d) 1p ≤ 335. If 1sin 10A = and 1sin ,5B = where A and B are positive actue angles, then A B+ is equal to (a) π (b) 2π (c) 3π (d) 4π 336. The value of sin1 ,sin1c° where 1c represents 1 rad, is (a) greater than 1 (b) less than 1 (c) equal to 1 (d) equal to /180π 337. The value of ( ) ( )sin sin 120 sin 240θ θ θ+ + ° + + ° is equal to (a) 0 (b) 1 (c) 3 (d) 2 338. The measure of the angle 114 35 30′ ′′° in radian is ………… (a) 1 rad (b) 2 rad (c) 3 rad (d) 4 rad 339. If ( ) ( )sin sinX A B A B= + − and ( ) ( )cos cos ,Y A B A B= + − then which one of the following is not correct? (a) 0,X Y′ ′+ > if 0 45B° < < ° for any A (b) 0,X Y+ = if 45B = ° for any A (c) X Y+ is a rational number of any A and B (d) 0,X Y+ < if 45 90B° < ≤ ° for any A 340. What is the value of tan 12π

? (a) 2 3− (b) 2 3+ (c) 2 3− (d) 3 2− 341. Which one of the following pairs is not correctly matched ? (a) sin 2π : ( )sin 2π− (b) tan 45° : ( )tan 315− ° (c) ( )1cot tan 0.5− : ( )1tan cot 0.5− (d) tan 420° : ( )tan 60− ° 342. If sec cos , sec sin , tan ,x a y b z cθ φ θ φ θ= = = then what is the value of 2 2 2

2 2 2 ?x y z

a b c+ −

(a) 1 (b) 0 (c) 1− (d) 2 2 2a b c+ − 343. If ,2A B

π+ = The greatest and the least values of cos cos ,A B are respectively ……….

(a) 12 and 0 (b) 0 and 1/ 2− (c) 12 and 12− (d) 0 and 1− 344. Given that tan tanp α β= + and cot cot ;q α β= + then what is the value of 1 1

p q

−

?

(a) ( )cot α β− (b) ( )tan α β− (c) ( )tan α β+ (d) ( )cot α β+



345. What is the value of ( ) ( ){ } ( )( ) ( ) ( ){ }

22cosec cot 9 / 2 cosec 2cot 2 sec sec 3 / 2π θ π θ π θ

π θ π θ π θ

+ − −

− − +

(a) 0 (b) 1 (c) 1− (d) ∞ 346. The value of sin 20 sin 40 sin 60° ° ° sin 80° is equal to (a) 3 /16− (b) 5 /16 (c) 3 /16 (d) 5 /16− 347. The value of ( )4 433 sin sin 32π α π α

− + +

( )6 62 sin sin 52π α π α − + + −

is equal to

(a) 0 (b) 1 (c) 3 (d) sin 4 sin 6α α+ 348. What is the value of tan15 tan195° ⋅ ° ? (a) 7 4 3− (b) 7 4 3+ (c) 7 2 3+ (d) 7 6 3+ 349. If 2 2 2 2 ....... cosec ,θ+ + + + ∞ = then the value of sinθ is equal to (a) 1 (b) 14 (c) 12 (d) 12 350. Which one of the following statements is correct ? (a) The squares of the tangents of the angles 30 , 45 , 60° ° ° are in GP (b) The squares of the sines of the angles 30 , 45 , 60° ° ° are in GP (c) The squares of the secants of the angles 30 , 45 , 60° ° ° are in AP (d) The squares of the tangents of the angles 30 , 45 , 60° ° ° are in AP 351. Consider the following statements I. If 1200θ = ° the ( ) 1sec tanθ θ −

+ is positive II. If 120 ,θ = ° then ( )cosec -cotθ θ is negative Which of the statements given above is/are correct ? (a) Only I (b) Only II (c) Both I and II (d) Neither I hor II 352. Match List I with List II and select the correct answer using the code given below the lists List I List II A. tan15° 1. 2 3− − B. tan 75° 2. 2 3+ C. tan105° 3. 2 3− Codes A B C A B C A B C A B C (a) 4 1 2 (b) 4 2 1 (c) 3 2 1 (d) 2 1 4 353. If 18 ,θ = ° then the value of 24sin 2sinθ θ+ is ………… (a) 1− (b) 1 (c) 0 (d) 2 354. The correct sequence of the following values is ……….. I. sin 12π

II. cos 12π

III. cot 12π

Select the correct answer using the code given below (a) III II I> > (b) I > II > III (c) I > III > II (d) III > I > II 355. Consider the following statements I. 1° in radan measure is less than .0.02 radians



II. I radian in degree measure is greater than 45 .° Which of the above statements is/are correct? (a) Only I (b) Only II (c) Both I and II (d) Neither I nor II 356. Let A and B be obtuse angles such that sin 4 / 5A = and cos 12 /13.B = − What is the value of ( )sin ?A B+ (a) 63 / 65− (b) 33 / 65− (c) 33 / 65 (d) 63 / 65 357. If cos 2a = and sin 7,b = then (a) 0, 0a b> > (b) 0ab < (c) a b> (d) a b<

358. If 3 ,2 2π πθ< < then 1 sin1 sinθθ−

+ is equal to

(a) sec tanθ θ− (b) sec tanθ θ+ (c) tan secθ θ− (d) None of these 359. If 0 ,2πθ< < and if 1 1 sin ,1 1 siny

y

θθ

+ +=

− − then y is equal to

(a) cot 2θ (b) tan 2θ (c) cot tan2 2θ θ+ (d) cot tan2 2θ θ

− 360. The set of all possible values of α in [ ],π π− such that 1 sin1 sinαα−

+ is equal to sec tan ,α α− is

(a) [ )0, / 2π (b) [ ) ( ]0, / 2 / 2,π π π∪ (c) [ ], 0π− (d) ( )/ 2, / 2π π− 361. If 3sin cosa x x= and 3cos sinb x x= , then (a) 0a b− > for ( )0, / 4x π∈ (b) 0a b− < for ( )0, / 4x π∈ (c) 0a b+ < for ( )0, / 2x π∈ (d) 0a b+ < for ( )0, / 4x π∈ 362. If A is an obtuse angle, then 3 3

2sin cos sin 2 tan cotsin cos 1 tanA A AA A

A A A

−+ −

− + is always equal to

(a) 1 (b) 1− (c) 2 (d) none of these 363. If tan 4 / 3,θ = − then sinθ is (a) 4 / 5− but not 4 / 5 (b) 4 / 5− or 4 / 5 (c) 4 / 5 but not 4 / 5− (d) none of these 364. ]If ( )

( )sin 'sin x y a b

x y a b

+ +=

− − then tantan xy is equal to

(a) ba

(b) ab

(c) ab (d) none of these 365. If tan cos sin ,2x ecx x= − then the value of 2tan ,2x is (a) 2 5− (b) 2 5+ (c) 2 5− − (d) 2 5− + 366. If 3cos '4A = then 5 532sin sin sin2 2 2A A A =

(a) 7 (b) 8 (c) 11 (d) none of these



367. If sin 2 cos3θ θ= and θ is an acute angle, then sinθ equals (a) 5 14− (b) 5 14 −

−

(c) 5 14+ (d) 5 14− − 368. The value of 2 4 8 14cos cos cos cos15 15 15 15π π π π is (a) 1 (b) 1/2 (c) 1/4 (d) 1/16 369. The value of 2 3 4 5 6 7cos cos cos cos cos cos cos15 15 15 15 15 15 15π π π π π π π is (a) 612 (b) 712 (c) 812 (d) None of these 370. The value of tan 5θ is (a) 3 5

2 45 tan 10 tan tan1 10 tan 5 tanθ θ θθ θ

− +− +

(b) 3 52 45 tan 10 tan tan1 10 tan 5 tanθ θ θθ θ

+ −− −

(c) 5 3

2 45 tan 10 tan tan1 10 tan 5 tanθ θ θθ θ

− −− +

(d) none of these 371. If 3 ,2ππ α< < then the expression 4 2 24sin sin 2 4cos 4 2π α

α α + + −

is equal to (a) 2 4sinα+ (b) 2 4sinα− (c) 2 (d) none of these 372. If α is an acute angle and 1sin ,2 2x xα −

= then tanα is (a) 11xx −+ (b) 11xx −+ (c) 2 1x − (d) 2 1x + 373. If sin cosA B m+ = and 3 3sin cos ,A A n+ = then (a) 3 3 0m m n− + = (b) 3 3 2 0n n m− + = (c) 3 3 2 0m m n− + = (d) 3 3 2 0m m n+ + = 374. If 3 ,2A B C

π+ + = then cos 2 cos 2 cos 2A B C+ + =

(a) 1 4cos cos cosA B C− (b) 4sin sin sinA B C (c) 1 2cos cos cosA B C+ (d) 1 4sin sin sinA B C− 375. The maximum value of ( ) ( )sin / 6 cos / 6x xπ π+ + + in the interval ( )0, / 2π is attained at (a) /12π (b) / 6π (c) / 3π (d) / 2π 376. If 2 ,π θ π< < then 1 cos1 cosθθ+

− is equal to

(a) cos cotecθ θ+ (b) cos cotecθ θ− (c) cos cotecθ θ− + (d) cos cotecθ θ− − 377. If ,2π θ π< < then 1 sin 1 sin1 sin 1 sinθ θ

θ θ− +

++ −

is equal to (a) 2secθ (b) 2secθ− (c) secθ (d) secθ−



378. If ( ) ( ){ } 1tan tan cos sec 1 1.cot cotα βα β α β

α β−+

+ − + + =+

then tan tanα β is equal to (a) 1 (b) 1− (c) 2 (d) 2− 379. The value of tan 70 tan 20tan 50° − °

=°

(a) 2 (b) 1 (c) 0 (d) 3 380. The minimum value of 1 '3sin 4cos 7θ θ− +

is (a) 112 (b) 512 (c) 712 (d) 16 381. The maximum value of cos 1 sincos '1 sin cosx x

xx x

− + −

is (a) 1 (b) 3 (c) 2 (d) 4 382. If tan .bx

a= then a b a b

a b a b

+ −+ =

− + where 0, 4x

π =

(a) 2sinsin 2xx (b) 2coscos 2xx (c) 2cossin 2xx (d) 2sincos 2xx 383. Let 0 / 4,x π< ≤ then ( )sec 2 tan 2x x− equals (a) ( )2tan / 4x π+ (b) ( )tan / 4x π+ (c) ( )tan / 4 xπ − (d) ( )tan / 4x π− 384. If cos sin ,

a b

θ θ= then sec 2 cos 2a b

ecθ θ+ is equal to

(a) a (b) b (c) ab

(d) a b+ 385. If ,α β γ π+ − = then 2 2 2sin sin sinα β γ+ − is equal to (a) 2sin sin sinα β γ (b) 2cos cos cosα β γ (c) 2sin sin sinα β γ (d) none of these 386. If tan cot ,4 4απ βπ =

then

(a) 0α β+ = (b) 2nα β+ = (c) 2 1nα β+ = + (d) ( )2 2 1 , .n n Zα β+ = + ∈ 387. If sincosA n

B= and sin ,cosA m

B= then ( )2 2 2sinm n B− =

(a) 21 n− (b) 21 n+ (c) 1 n− (d) 1 n+ 388. If ( ) ( )cos cos ,mθ φ θ φ+ = − then tanθ is equal to (a) 1 tan1 m

mφ

+−

(b) 1 tan1 m

mφ

−+

(c) 1 cot1 m

mφ

−+

(d) 1 sec1 m

mφ

+−

389. If coscos x a

y= and sin ,sin x b

y= then : ( )2 2 2sin ...a b y− ⋅ =



(a) 2 1a − (b) 2 1b − (c) 2 1x − (d) 2 1y − 390. If 33 2ππ θ< < , then 2 2 2cos 4θ+ + is equal to (a) 2cosθ− (b) 2sinθ− (c) 2cosθ (d) 2sinθ 391. 3cos cos33sin sin 3θ θ

θ θ+−

is equal to (a) 21 cot θ+ (b) 4cot θ (c) 3cot θ (d) 2cotθ 392. 2 23 4cos cos5 5π π

+ is equal to (a) 45 (b) 52 (c) 54 (d) 34 393. If 3sin 5sinα β= , then tan 2tan 2

α β

α β

+

− is equal to

(a) 1 (b) 2 (c) 3 (d) 4 Directions (Q. Now. 394-396) Let ( )sin 1A B+ = and ( ) 1sin ,2A B− = where , 0, 2A B

π ∈

394. What is the value of A? (a) 6π (b) 3π (c) 4π (d) 8π 395. What is the value of ( ) ( )tan 2 tan 2 ?A B A B+ + (a) 1− (b) 0 (c) 1 (d) 2 396. What is the value of 2 2sin sin ?A B− (a) 0 (b) 12 (c) 1 (d) 2


120cm θ

O

Only one option is correct. 1. Ans. (a), ∵ π radians 180= ° , ∴ 1 radian o180

π =

o o180 7 63022 11× = =

57 17 45′ ′′≈ ° 2. Ans. (c), Given, 34 radian is the angle subtended at the centre by arc of length 15 cm ∴ 2π radians is the angle subtended by arc of length 15 23/ 4 π = ×

cm 40π= cm

∴ Circumference of the circle 40π= cm 2 40rπ π⇒ = where r is the radius of the circle 402r

ππ

⇒ =

cm 20= cm 3. Ans. (c), Length of the arc 180rl

π θ= [Where θ is in degrees]

22 128 457 360 = × × ×

cm 22= cm 4. Ans. (c), Let θ be the angle subtended at the centre. ∴ Radius of the circle 502r = cm = 25 cm, Length of the arc 11l = cm Now, 180rl

π θ= , where θ is in degrees o o180 11 180 722 25l

rθ

π× × × ⇒ = = ×

o126 25 125 ′= = °

5. Ans. (a), At the time 4 : 20 , the minutes hand is exactly at 4. The hour hand is at 4 at time 4 : 00. The hour hand traces an angle of ( )2 cπ in 12 hours, i.e. it traces an angle 212

cπ

in 1 hour, i.e., in 60 min. So, the angle traced by the hour hand in 20 min 2 2012 60 18cπ π = × =

radians.

∴ The angle between the minute hand and the hour hand at time 4 : 20 is 18π radians. 6. Ans. (d), Let the three angles of the triangle be ,4 4π π

θ+ and 24π θ+ . Then, 24 4 4π π π

θ θ π + + + + =

[∵ Sum of angles of a triangle π= ] 3 34π θ π⇒ + = 3 4πθ⇒ = 12πθ⇒ = ∴ The largest angle 524 12 12π π π = + =

.

7. Ans. (b), Length of the wire = circumference of a circle of radius 15 cm 2 15 30 cmπ π= × = ∴ Length of the arc formed by the wire when placed on the loop 30 cmπ= Now, the length of arc of a circle of radius r , subtending an angle θ at the circle is given by 180rS

π θ= where θ is in degrees

( )2 12030 360π θπ

× ×⇒ = 45θ⇒ = °

SSOOLLUUTTIIOONN OOFF TTRRIIGGOONNOOMMEETTRRIICC FFUUNNCCTTIIOONN ((�))

JEE (MAIN & ADV.), MEDICAL + BOARD, NDA, X & IX Enjoys unparalleled reputation for best results in terms of percentage selection www.newtonclasses.net



108°132 m

OA B

C

XY

r r1θ 2θ

C′

B′Y

A′ X

80°••4π/9C

AO

B

∴ Angle subtended at the circle 45= ° 8. Ans. (b), Let the polygon have n sides. Then, the sum of its interior angles = Sum to n terms of an A.P. with first term 120° and common difference 5° ( ) ( ){ }

o2 120 1 52n n = × + − ⋅

( )o5120 12n n n

= + −

Also, sum of the interior angles ( )2 4 90n= − × ° [∵ Sum of the interior angles of a polygon of n sides ( )2 4 90n= − × ° ] ∴ ( ) ( )52 4 90 120 12n n n n− × = + − 2360 720 240 5 5n n n n⇒ − = + − 25 125 720 0n n⇒ − + = 2 25 144 0n n⇒ − + = ( )( )16 9 0n n⇒ − − = 9n⇒ = or 16n = But when 16n = , then the largest angle is 195° which is more than 180° . Hence, 16n ≠ ∴ The polygon has 9 sides. 9. Ans. (c), The length ‘ l ’ of the arc of a circle of radius r , subtending an angle

θ at the centre is given by, 2360rlπ θ

= where θ is in degrees 22 108132 2 7 360r⇒ = × × × 132 7 3602 22 108r

× ×⇒ =

× × 70= m.

10. Ans. (a), We have 42m A∠ = ° and c 180 306 6o

m Bπ π

π ∠ = = × = °

But, in ,ABC∆ 180m A m B m C∠ + ∠ + ∠ = ° ∴ 42 30 180m C° + ° + ∠ = ° ∴ 108m C∠ = °

c108180π = ×

c35π =

Thus, 3108 5

c

m Cπ ∠ = ° =

11. Ans. (b), We are given 12

45θθ

= …(i) ∴ Area of sectorArea of sectorOAXCOBYC

2 12 2

1212r

r

θ

θ= 1

245θ

θ= = [Using (i)]

Clearly, ratio of areas = ratio of angles 4 : 5= 12. Ans. (c), From the figure 18 cmOA = and 36OA′ = cm ∴ 36 18 18AA OA OA′ ′= − = − = cm similarly, 18BB′ = cm ∴ 18AA BB′ ′= = cm Then, 80m AOB m A OBθ ′ ′= ∠ = ∠ = ° c c480180 9π π = × =

∴ Length of ( )arc ACB ( )arcl A C B′ ′ ′= ( )( ) 418 89OAπ

θ π= = × = cm and length of ( )arc A C B′ ′ ′ ( )arcl A C B′ ′ ′= ( )( ) 436 16 cm9OA

πθ π′= = × =

∴ Perimeter of required region AA C B BCA′ ′ ′ ( ) ( )AA BB l ACB l A C B′ ′ ′ ′ ′= + + +



6 5 439

12

78211110 HH MH

18 18 8 16π π= + + + ( )36 24 6 6 4π π= + = + cm Further, Area of ( )sector OACB = ( ) ( )21sector 2A OACB OA θ= ⋅ ( )21 4182 9π= × × 272 cmπ= Area of ( )sector OA C B′ ′ ′ = ( ) ( )21sector 2A OA C B OA θ′ ′ ′ ′= ⋅ ( )2 4362 9π1

= × × 2216 cmπ= ∴ Area of shaded region ( ) ( )A OA C B A OACB′ ′ ′= − 288 72π π= − 2216 cm= 13. Ans. (a), When the time is 2 : 20 : 00 (twenty minutes past two), the minute-hand is exactly at mark 4 and the hour-hand has crossed one-third of the angle between mark 2 and mark 3 Now, the hour-hand moves through one complete rotation ( )360° in 12 hours ∴ In 1 hour, the hour-hand moves through ( )360 /12 ° ∴ In 60 minutes, the hour-hand moves through 30° ∴ In 20 minutes, the hour-hand moves through 20 30 ,30

o

×

i.e. 10° ∴ angle between hour-hand and mark 3 is 30 10 20° − ° = ° Also, angle between mark 3 and mark 4 is 30° ∴ Angle between the hour-hand and minute-hand at 2 : 20 is 20 30 50° + ° = ° 14. Ans. (b), Angle between the hands at 3 is 90° . In 20 minutes, minute hand traces an angle

6 20 120°× = ° , and hour hand traces an angle o1 20 102 × = ° ∴ Angle between the hands 120 90 10 20 9π= °− ° − ° = ° = 15. Ans. (b), Let the angles be ,A B and C ∴ Given information ⇒ : : 1: 2 :3A B C = Therefore, , 2A k B k= = , 3C k= and A B C π+ + = 2 3k k k π⇒ + + = ⇒ 6k

π= ∴ Angles are , and 6 3 2π π π , Clearly the smallest angle is 6π

16. Ans. (d), Perimeter of the circle of radius 5 cm is 2 .5 10π π= cm = Arc length of the new circle (10π cm) Given, radius r of the new circle = 10 cm. Now we know that 10 10s rθ π θ= ⇒ = θ π⇒ = 17. Ans. (c), Given, length of wire is 40π cm, we know that ( )40 1 100 40 100s rθ π θ π θ= ⇒ = × ⇒ = 25πθ⇒ = 18. Ans. (c), As area of a circle is 2rπ ∴ 29 rπ π= 3r⇒ = , Hence, perimeter of the sector 2 2 3 3 60 6 3 6180 3r r

π πθ π = + = × + × °× = + = + °

19. Ans. (b), As three terms in A.P. are , ,a d a a d− + So, angles be ,a d a− and a d+ , As we know that sum of the three angles of a triangle 180= ° ⇒ 180a d a a d− + + + = ° 60a⇒ = ° , But given ( )2a d a d+ = − 3ad⇒ = 20d⇒ = °



∴ The angles are 60 20 , 60 , 60 20°− ° ° ° + ° 40 , 60 , 80= ° ° ° , Hence, the option (b) is correct. Alternatively : If angles of a triangle are in A.P., then one of the angles must be 60° . ∴ Option (c) is not a solution Option (d) is not a solution because their sum is not equal to 180° . In option (a), the angles are in A.P, but greatest angle ≠ double the least angle. ∴ Option (b) is the required solution. 20. Ans. (c), The wheel makes 120 revolutions in 60 seconds. ∴ In 1 second wheel makes 2 revolutions, and in 2 seconds it makes 4 revolutions. Also in one revolution second wheel max 2π radian, ∴ Angle covered in 2 seconds 3 2 8π π= × = 21. Ans. (b), Let the two polygons A and B have 3x and 4x sides respectively. An interior angle of polygon 2 3 4 6 490 903 3x x

Ax x

× − −= × ° = × °

and an interior angle of polygon 2 4 4 8 490 904 4x xB

x x

× − −= × ° = × °

∴ 8 4 6 490 90 154 3x x

x x

− − × ° − × ° = °

[Note : The polygon with larger no. of sides has larger interior angle] 8 4 6 4 154 3 90x x

x x

− − ° ⇒ − = ° 24 12 24 16 112 6x x

x

− − +⇒ = 4 112 6x

⇒ = 2x⇒ = . ∴ The two polygons have 6 and 8 sides respectively. ∴ The polygon with larger interior angle has 8 sides. ( )Note that sum of all interior angles of any polygon of sides 2 4 2n n

π = −

22. Ans. (a), Since 0,2 Aπ

− < < we have sin 0 i.e., sin is veA A< − Since 02 B

π− < < , we have sin 0 i.e., sin is veB B< −

Now, cos cos 13 4 5A B= = 3 4cos and cos .5 5A B⇒ = =

2 9 4sin 1 cos 1 25 5A A∴ = − − = − − = − and 2 16 3sin 1 cos 1 25 5B B= − − = − − = − 4 32sin 4sin 2 4 4.5 5A B

∴ + = − + − = −

23. Ans. (b), ∵ sin cos tan 1sin cos tan 1θ θ θ

θ θ θ+ +

=− −

[Dividing Nr. and Dr. by cosθ ] 1

1p

p qq

p p q

q

++

= =−−



24. Ans. (c), cos9 sin 9 1 tan 9cos9 sin 9 1 tan 9° + ° + °=

° − ° − ° [Dividing Nr. And Dr. by cos9° ]

( )tan 45 tan 9 tan 45 9 tan 54 .1 tan 45 . tan 9° + °= = °+ ° = °

− ° ° [ ]tan 45 1° =∵ and ( ) tan tantan 1 tan tanA B

A BA B

+ + = − ⋅ ∵

25. Ans. (d), Given, cos sin 1θ θ− = cos 1 sinθ θ⇒ = + Clearly, sin 0θ = When 0 , then sin is ve and so 1 sin 1But since cos 1, so we have sin 0θ π θ θ

θ θ< ≤ + + ≥

≤ =

∵ But for cos 1θ π θ= = − , Hence no solution 26. Ans. (b), cos1 .cos 2 ........cos100° ° ° cos1 .cos 2 ........cos89 .cos90 .cos91 .....cos100= ° ° ° ° ° ° 0= [ ]cos90 0° =∵ 27. Ans. (d), ( )22 2cos sec cos sec 2cos secp θ θ θ θ θ θ= + = − + ( )2cos sec 2 2θ θ= − + ≥ , ∴ 2p ≥ [ ]cos sec 1θ θ⋅ =∵ 28. Ans. (d), ( ) ( ) ( )cos 270 cos 90 sin 270 cosθ θ θ θ°+ ° − − °− ( ) 2 2sin .sin cos cos sin cos 1θ θ θ θ θ θ= − − = + = 29. Ans. (b), cos1 .cos 2 ....cos179° ° ° cos1 . cos 2 ......cos89 . cos90 . cos91 .......cos179= ° ° ° ° ° ° 0= [ ]cos90 0° =∵ 30. Ans. (a), tan 1 sec1 sec tanA A

A A

++

+

( )

2 2tan 1 sec 2sectan 1 secA A A

A A

+ + +=

+

( )

22sec 2sectan 1 secA A

A A

+=

+

( )( )

2sec 1 sec 2sec 2cosectan 1 sec tanA A AA

A A A

+= = =

+

31. Ans. (c), ∵ 4 4cos sinθ θ− ( ) ( )2 22 2cos sinθ θ= − ( )( ) ( )2 2 2 2 2 2cos sin cos sin cos sinθ θ θ θ θ θ= − + = − ( ){ }2 2 2cos 1 cos 2cos 1θ θ θ= − − = − 32. Ans. (d), 52sin sin12 12π π

2sin15 sin 75 2sin15 cos15= ° ° = ° ° sin 30= ° 12= 33. Ans. (a), Since α is a root of 225cos 5cos 12 0θ θ+ − = , so we have 225cos 5cos 12 0α α+ − = 5 25 1200 5 35cos 50 50α

− ± + − ±⇒ = = 4cos 5α⇒ = − cos is -ve2π α π α < < ⇒

∵ ∴ 16 3sin 1 25 25α = − = sin is +ve2π α π α < < ⇒

∵ and so, 3 4 24sin 2 2sin cos 2 .5 5 25α α α = = × × − = −

34. Ans. (d), ( ) 1sin 2x y− = 30 or150x y⇒ − = ° ° and ( ) 1cos 2x y+ = 60 or 300x y⇒ + = ° ° Now, we have (i) When 30 , 60 ,x y x y− = ° + = ° then 45 , 15x y= ° = ° (ii) When 150 , 60 ,x y x y− = ° + = ° then 105 , 45x y= ° = − ° (iii) When 30 , 300 ,x y x y− = ° + = ° then 165 , 135x y= ° = °



(iv) When 150 , 300 ,x y x y− = ° + = ° then 225 , 75x y= ° = ° Since andx y lie between 0 and180° ° we have, ( ) ( )45 , 15 or 165 , 135x y x y= ° = ° = ° = ° Out of the given alternatives we have 45 , 15 .x y= ° = ° So, option (d) is correct 35. Ans. (b), 1 sin 1 sin1 sin 1 sinθ θ

θ θ− +

++ −

1 sin 1 sin 1 sin 1 sin1 sin 1 sin 1 sin 1 sinθ θ θ θθ θ θ θ

− − + + = × + + − − +

( ) ( )2

1 sin 1 sin1 sinθ θ

θ

− + +=

− 2 2 seccos θ

θ= =

But since θ lies in the second quadrant, secθ is negative and so sec secθ θ= − 1 sin 1 sin 2sec1 sin 1 sinθ θ

θθ θ

− +∴ + = −

+ −

36. Ans. (b), ( )( )

( )( )

222 1 31 tan 601 tan2 tan 2 tan 60 2 3θθ

++ °+= =

° 4 2 .2 3 3= =

Alternatively : 2 21 tan sec 1 12 tan 2 tan 2sin cos sin 2θ θ

θ θ θ θ θ+

= = = ( )

1 1 1 2sin120 sin 180 60 sin 60 3= = = =° °− ° °

37. Ans. (a), Given, sec mθ = and tan nθ =

( )1 1sec tanm n θ θ

∴ =+ +

( )

( )( )sec tan1 .sec tan sec tanθ θ

θ θ θ θ−

=+ −

( )( ) ( )2 2sec tan sec tansec tanθ θ

θ θθ θ

−= = −

− 2 2sec tan 1θ θ − = ∵

Now, ( )( )

1 1m n

m m n

+ +

+ ( ) ( ){ }1 1sec tan sec tan .2sec 2.sec secθ θ θ θ θ

θ θ= + + − = =

38. Ans. (c) 1 sin cos sin (1 cos )1 sin cos sin (1 cos )θ θ θ θθ θ θ θ

+ − + −=

+ + + +

22

2sin cos 2sin2 2 22sin cos 2cos2 2 2θ θ θ

θ θ θ

+=

+ = 2sin sin cos2 2 2 tan 22cos sin cos2 2 2

θ θ θθ

θ θ θ

+ = +

. 39. Ans. (c), Given, sin cosx A A= + sin130 cos130= °+ ° ( ) ( )sin 180 50 cos 90 40= °− ° + °+ ° sin 50 sin 40 0= °− ° > [ ]sin 50 sin 40° > °∵ [Note: In the 1st quadrant sinθ is always increasing] 40. Ans. (b), Given, 3cos 2α = − 3 1 1sin 1 4 4 2α⇒ = − = = sin is +ve when 2πα α π < <

∵ sin 1/ 2 1tan cos 3 / 2 3α

αα

∴ = = = −−

1and cot 3tanαα

= = − and 3sin 5β = − 9 16 4cos 1 25 25 5β⇒ = − − = − = − 3cos is -ve when 2πβ π β < <

∵



and sin 3 / 5 3tan cos 4 / 5 4ββ

β−

= = =−

( )

2 21 33. 2.3 tan 2 tan 43 4cot cos 3 5

α βα β

− + + ∴ =+ − + −

31 52 .4 223 5− +

= =−

41. Ans. (b), Given, 1tan 7θ = 2 2 1 8sec 1 tan 1 .7 7θ θ⇒ = + = + = Also, cos 7θ = 2 2cosec 1 cot 1 7 8θ θ⇒ = + = + = , 2 2

2 288cosec sec 48 378cosec sec 64 48 7

θ θθ θ

−−∴ = = =

+ +

Alternatively : 2 2 2 22 2

2 2 2 22 2

1 1cosec sec cos sinsin cos1 1cosec sec cos sinsin cosθ θ θ θθ θθ θ θ θ

θ θ

−− −= =

+ ++ 2

21 tan1 tan θθ

−=

+ [Dividing Nr. and Dr. by 2cos θ ]

11 6 37 .1 8 41 7−

= = =+

42. Ans. (d), 3 tan 4 0A+ = 4tan 3A⇒ = − , 1 3cot tan 4A

A∴ = = − ,

Also, 2 16 5sec 1 tan 1 9 3A A= − + = − + = − [Note : sec A is ve− since A lies in 2nd quadrant] 1 3cos sec 5A

A= = − and 2 9 4sin 1 cos 1 25 5A A= − = − =

[Note : sin A is +ve since A lies in 2nd quadrant] Now, 2cot 5cos sinA A A− + 3 3 4 3 4 232 5 34 5 5 2 5 10 = − − − + = − + + =

43. Ans. (b), ∵ ( ) tan tantan 1 tan tanα βα β

α β+

+ =−

11 2 111 1 2 1x

x xx

x x

++ +=

− + +

( ) ( )( )( )

( )( )( )

2222

2 11 2 1 2 2 1 12 2 12 3 11 2 1

x x x

x x x x

x xx x x

x x

+ + +

+ + + + = = =+ + + + −

+ +

4πα β∴ + =

44. Ans. (c), cot 54 tan 20tan 36 cot 70° °+

° °( )

( )cot 90 36 tan 20tan 36 cot 90 20°− ° °

= +° °− °

( )( )

cot 90 36 tan 20tan 36 cot 90 20°− ° °= +

° °− °

tan 36 tan 20tan 36 tan 20° °= +

° ° 1 1 2= + = ( )cot tan 90θ θ= °− ∵



45. Ans. (b), Given, 5tan 6α = and 1tan 11β = , ∵ ( )5 1tan tan 6 11tan 5 11 tan tan 1 6 11

α βα β

α β

+++ = =

− − ×

6166 1 tan61 466π

= = =

, 4πα β∴ + = 46. Ans. (c), ∵ ( ) tan 70 tan 20tan 50 tan 70 20 1 tan 70 tan 20°− °

° = ° − ° =+ ° °

tan 70 tan 20 1 tan 70 tan 20tan 50° − °

⇒ = + ° °°

1 tan 70 cot 70= + ° ° ( )cot tan 90θ θ= °− ∵ 1 1 2= + = [ ]tan .cot 1θ θ =∵ 47. Ans. (d),. ∵ ( ) ( )tan 945 tan 945− ° = − ° ( )tan tanθ θ− = − ∵ ( )tan 1080 135= − °− ° ( )tan 3 360 135= − × °− ° tan135= ° ( )tan 6 tanπ θ θ− = − ∵ ( )tan 180 45= °− ° tan 45= − ° 1= − ( )tan tanπ θ θ− = − ∵ 48. Ans. (a) Clearly, the angle 200° lies in the III rd quadrant and so both sin 200° and cos 200° are negative. sin 200 cos 200∴ °+ ° is negative. 49. Ans. (b), We have: 180 and 180A C B D+ = ° + = ° [∵ The opposite angles of a cyclic quadrilateral are supplementary] cos cos cos cosA B C D∴ + + + ( ) ( )cos cos cos 180 cos 180A B A B= + + °− + °− cos cos cos cos 0.A B A B= + − − = 50. Ans. (d), ( ) ( )

( ) ( )tan 270 15 tan 360 15tan 255 tan 345tan195 tan105 tan 180 15 tan 90 15°− ° + ° − °°+ °

=°− ° ° + ° − ° + °

2

2cot15 tan15 cot 15 1tan15 cot15 1 cot 15°− ° ° −= =

°+ ° + ° [Dividing Nr. & Dr. by tan15° ]

22 11m

m

−=

+ .

51. Ans. (b), It is given that 5cosec cot 2θ θ+ = …(i) 2cosec cot 5θ θ− = …(ii) 1 cosec cotcosec cot θ θ

θ θ = + − ∵

Subtracting (ii) from (i) we get, 5 2 212cot 2 5 10θ = − = 21cot 20θ⇒ = 20tan .21θ⇒ = 52. Ans. (b), ( ) ( )cos cos 90 sin sin 90θ θ θ θ°− − °− cos sin sin cos 0θ θ θ θ= − = ( ) ( )cos 90 sin & sin 90 cosθ θ θ θ° − = ° − = ∵ 53. Ans. (c), Given, 1cos 2θ = − , Since cosθ is –ve, so θ lies in II or III quadrant.



Also, since 1cos 60 ,2° = so we have, ( ) 1cos 180 60 2° − ° = − and ( ) 1cos 180 60 2° + ° = − i.e., 1 1cos120 and cos 2402 2° = − ° = − i.e., 120 or 240θ = ° ° 54. Ans. (d), 2sin 1 cos sin1 1 cos sin 1 cosθ θ θ

θ θ θ+

− + −+ −

( )( )

( ) ( )( )

2 21 cos 1 cos 1 cos sin1 1 cos sin 1 cosθ θ θ θθ θ θ

− + − −= − +

+ −

( )( )

( )( )( )

2 2 21 cos 1 cos sin1 1 cos sin 1 cosθ θ θ

θ θ θ

− − −= − +

+ − ( )

( )( )

2 2sin sin1 1 cos cos 0 cos .sin 1 cosθ θθ θ θ

θ θ−

= − − + = + =−

55. Ans. (b), 6 6 2 2sin cos 3sin cosθ θ θ θ+ + ( )6 6 2 2 2 2sin cos 3sin cos sin cosθ θ θ θ θ θ= + + + 2 2sin cos 1θ θ + = ∵ ( )32 2sin cosθ θ= + 31 1.= = 2 2sin cos 1θ θ + = ∵ 56. Ans. (c), cosec cot qθ θ− = …(i) Taking inverse an both dies we get, 1cosec cot

qθ θ+ = …(ii)

Subtracting (i) from (ii) we get, 12cot qq

θ = − 21cot .2qqθ−

⇒ = 57. Ans. (b), Given, sec 2A = 1cos 2A⇒ = thcos is ve since lies in IV quadrantA A + ∵ 1 1 1sin 1 2 2 2A

−∴ = − − = − = thsin is ve when lies in IV quadrantA A − ∵

sin 1/ 2tan 1cos 1/ 2AA

A

−∴ = = = − thtan is ve when lies in IV quadrantA A − ∵

1 1cot 1tan 1AA

= = = −−

thcos is ve when lies in IV quadrantA A − ∵ and 1 1cosec 2sin 1/ 2A

A= = = −

− thcosec is ve when lies in IV quadrantA A − ∵

Now, ( ) ( )( ) ( )

1 1 21 tan cosec1 cot cosec 1 1 2A A

A A

+ − + −+ +=

+ − + − − − 2 1.2−= = −

58. Ans. (d), Given, 5 tan 4θ = 4tan 5θ⇒ = Now, 5sin 3cossin 2cosθ θ

θ θ−+

5 tan 3tan 2θθ−

=+

[Dividing Nr. & Dr. by cosθ ] 45 3 1 55 .4 14 1425 5

× − = = = +

59. Ans. (b), ∵ 1cot ,3 3π

= cot 1and cot 3.4 6π π= =



Clearly, 1cot . cot . 33 6 3π π =

21 cot 4π = =

cot , cot and cot are in G.P.3 4 6π π π∴

60. Ans. (c), 333cos cos3 3cos 4cos 3cos3sin sin 3 3sin 3sin 4sinθ θ θ θ θ

θ θ θ θ θ+ + −

=− − +

3 334cos cot4sin θθ

θ= = .

61. Ans. (a), 3 3 3log tan1 log tan 2 ..... log tan89° + ° + + ° ( ) ( )3 3 3 3log tan1 log tan89 log tan 2 log tan88 ....= °+ ° + °+ ° + ( )3 3 3log tan 44 log tan 46 log tan 45+ °+ ° + ° ( ) ( ) ( )3 3 3 3log tan1 . tan89 log tan 2 . tan 88 ..... log tan 44 . tan 46 log tan 45= ° ° + ° ° + + ° ° + ° ( ) ( )

( )3 33 3

log tan1 . cot1 log tan 2 cot 2 .....log tan 44 . cot 44 log tan 45= ° ° + ° ° +

+ ° ° + ° ( )tan cot 90θ θ= °− ∵

( )3 3 3 3log 1 log 1 .... log 1 log 1= + + + + [ ]tan .cot 1and tan 45 1θ θ = ° =∵ ( )0 0 ... 0 0 0.= + + + + = 62. Ans. (c), We have c c1 57 i.e. 1 1≈ ° > ° Also sinθ being an increasing function we have , sin1 sin1 i.e. sin1 sin1c > ° ° < 63. Ans. (b), tan1 . tan 2 . tan 3 ..... tan89° ° ° ° ( ) ( ) ( ) ( )tan1 . tan 89 . tan 2 . tan88 . tan 3 . tan 87 .... tan 44 . tan 46 .tan 45° ° ° ° ° ° ° ° ° ( ) ( ) ( ) ( )tan1 . cot1 . tan 2 cot 2 . tan 3 .cot 3 .... tan 44 . cot 44 . tan 45= ° ° ° ° ° ° ° ° ° ( )tan 90 cotθ θ° − = ∵ 1.1.1...1.1= [ ]tan .cot 1and tan 45 1θ θ = ° =∵ 1.= 64. Ans. (b), 2 144 5cos 1 sin 1 169 13θ θ= − = − = stNote : cos is ve since lies in 1 quadrantθ θ + 2 9 4sin 1 cos 1 25 5φ φ= − − = − − = − ndNote: sin is ve since lies in 2 quadrantφ φ − Now, ( )sin sin cos cos sinθ φ θ φ θ φ+ = + 12 3 5 4 56. . .13 5 13 5 65 = − + − = −

65. Ans. (b), 221 tan 151 tan 15− °

+ ° ( )cos 2 15= × ° 3cos30 .2= ° =

66. Ans. (d), Given, 1 2tan cotkθ θ= 1 2tan tan kθ θ⇒ ⋅ = Hence ( )

( )1 2 1 2 1 21 2 1 2 1 2

cos cos cos sin sincos cos cos sin sinθ θ θ θ θ θθ θ θ θ θ θ− +

=+ −

1 21 2

1 tan tan1 tan tanθ θθ θ

+=

− [ ]1 2Dividing Nr.and Dr.by cos cosθ θ

1 .1 k

k

+=

− [ ]1 2 1 2tan cot tan tank kθ θ θ θ= ⇒ =∵

67. Ans. (a), ( ) ( )cos 480 . sin150 sin 600 . cos390 cos 360 120 .sin 180 30° ° + ° ° = °+ ° ° − ° ( ) ( )sin 540 60 .cos 360 30+ °+ ° °+ ° cos120 . sin 30 sin 60 . cos30= ° °− ° ° cos 60 . sin 30 sin 60 . cos30= − ° ° − ° ° ( )cos120 cos 180 60 cos60° = ° − ° = − ° ∵ ( )sin 60 .cos30 cos 60 .sin 30= − ° ° + ° ° ( )sin 60 30 sin 90 1.= − °+ ° = − ° = −



68. Ans. (a), Given, ( )2cos sin sin cos 1 0x x k x x+ + − = 2 2cos sin 2sin cos sin cos 1 0x x x x k x x⇒ + + + − = 1 2sin cos sin cos 1 0x x k x x⇒ + + − = 2 2cos sin 1x x + = ∵ ( )2 sin cos 0k x x⇒ + = which can be true for all x if 2k = − 69. Ans. (a), ∵ ( )cot105 cot 90 15° = ° + ° ( )tan15 tan 45 30= − ° = − °− ° tan 45 tan 301 tan 45 tan 30° − ° = − + ° °

11 3 131 3 11 1. 3

− −= − = −

+ +

( )( )( )( )

3 1 3 1 4 2 3 3 223 1 3 1− − −= − = − = −

+ −, ∴ cot105 3 2° = −

70. Ans. (a), ( ) ( ) ( )sin sin sinsin sin sin sin sin sinα β β γ γ αα β β γ γ α− − −

+ + sin cos cos sin sin cos cos sin sin cos cos sinsin sin sin sin sin sinα β α β β γ β γ γ α γ α

α β β γ γ α− − −

= + + ( ) ( ) ( )cot cot cot cot cot cot 0β α γ β α γ= − + − + − = 71. Ans. (a), Given, 2sin sin 1x x+ = 2 2sin 1 sin cosx x x⇒ = − = Squaring on both sides, we get 2 4sin cosx x= , ∴ 2 4 2cos cos sin sin 1x x x x+ = + = 72. Ans. (d), Given, cosax h

θ= + cosax h

θ⇒ − = cosa

x hθ⇒ =

− …(i)

and sinby kθ

= + sinby kθ

⇒ − = ⇒ sinb

y kθ=

− …(ii)

Squaring and adding (i) and (ii) we get, ( ) ( )

2 22 2 1a b

x h y k+ =

− −

73. Ans. (a), 1tantan cot tan1tan cot tan tanθ

θ ϕ ϕϕ θ ϕ

θ

−−

=− −

tan tan 1tantan tan 1tanθ ϕ

ϕθ ϕ

θ

−

=−

tan tan 1 tan tantan tan tan 1 tanθ ϕ θ θϕ θ ϕ ϕ

−= ⋅ =

−

74. Ans. (c), ∵ We know that ( )tan85 tan 90 5 cot 5° = °− ° = ° ∴ tan 5 tan85 tan 5 cot 85 1° ⋅ ° = °⋅ ° = Now, let us group the terms as ( ) ( ) ( ) ( )tan 5 tan85 tan15 tan 75 tan 25 tan 65 tan 35 tan 55 tan 45 1° ⋅ ° ⋅ ° ⋅ ° ⋅ ° ⋅ ° ⋅ ° ⋅ ° ⋅ ° = 75. Ans. (b), Given, 2 2 2 2cos 54 sin cos 36 sinθ θ= °− + °− 2 2 2 2sin 36 cos 36 2sin 1 2sin cos 2θ θ θ= °+ °− = − = Alternatively : ( ) ( ) ( ) ( )cos 54 cos 54 cos 36 cos 36θ θ θ θ°+ ° − + °− ⋅ ° + ( ) [ ] [ ] ( )cos 54 cos 90 90 54 cos 90 90 36 cos 36θ θ θ θ= °+ ⋅ ° − °+ °− + °− °+ ° − °+ ( ) ( ) ( ) ( )cos 54 sin 36 sin 54 cos 36θ θ θ θ= °+ ⋅ ° + + °+ ⋅ °+ ( ) ( )sin 54 36 sin 90 2 cos 2θ θ θ θ= °+ + °+ = °+ =



[As we know that sin cos cos sinA B A B+ ( )sin A B= + ] 76. Ans. (d), 3sin 5A = 2 9 4cos 1 sin 1 25 5A A⇒ = − − = − − = − and 2 144 5sin 1 cos 1 169 13B B= − − = − − = − , Hence sin 3tan cos 4A

AA

= = − and sin 5tan cos 12BB

B= =

Hence ( ) tan tantan 1 tan . tanA BA B

A B

++ =

− 3 5 9 54 12 1248 153 51 484 12

− +− +

= =+ − −

4 48 1612 63 63= − × = − Alternatively :

From figure it is clear that 3 3tan 4 4A = = −

− and 5 5tan 12 12B

−= =−

Therefore, ( ) tan tantan 1 tan . tanA B

A BA B

++ =

− 3 5 9 54 12 1248 153 51 484 12

− +− +

= =+ − −

4 48 1612 63 63= − × = −

77. Ans. (c) 22

2sin1 cos 2 tan1 cos 22cos 2x

θθ θ

θθ−

= = =+

Hence 2 22 tan2 2 tan 2 tan(1 ) 21 tan 2

x

x

θθ

θθ

= = × = − −.

78. Ans. (a), sin sin21 cos cos2θ

θ

θθ

+

+ + 2

sin 2sin cos2 2 21 cos 2cos 12 2θ θ θ

θ θ

+=

+ + − 2

sin 1 2cos2 2cos 2cos2 2θ θ

θ θ

+ =+

sin 1 2cos2 2 tan 2cos 1 2cos2 2θ θ

θθ θ

+ = = +

79. Ans. (a), 2

21 cos 2 sin 2 2cos 2sin cos1 cos 2 sin 2 2sin 2sin cosθ θ θ θ θθ θ θ θ θ

+ + +=

− + +( )( )

2cos cos sin cot2sin sin cosθ θ θθ

θ θ θ+

= =+

80. Ans. (b), 2 2 2cos4θ+ + ( )2 22 2 2 2cos 2 1 2 2 4cos 2 2θ θ= + + − = + + − { }22 2cos 2 2 1 2cos 1θ θ= + = + − [∵ 0, 4πθ ∈

2 0, 2πθ ⇒ ∈

,Hence 2cos 2 cos 2θ θ= ]

24cos 2 cosθ θ= = 2cosθ= stis in 1 quadrantθ ∵

O

Y

A

X

54−3 O

Y

B

X12−5− 13



81. Ans. (c), Given, ( )( )

sin 2 sin 5 sin sin 2 2cos3 sin 2cos 2 cos5 cos cos 2 2cos3 cos 2x x x x x x

x x x x x x

+ − +=

+ + + ( )

( )sin 2 1 2cos3 tan 2cos 2 1 2cos3x x

xx x

+= =

+

82. Ans. (b), ( ) ( )( ) ( )

sin 1 sin 1cos 1 2cos cos 1n n

n n n

α αα α α+ − −

+ + + −

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

1 1 1 12cos sin2 21 1 1 12cos cos 2cos2 2n n n n

n n n nn

α α α α

α α α αα

+ + − + − − ⋅

=+ + − + − −

⋅ +

2cos sin2cos .cos 2cosn

n n

α αα α α

=+

sincos 1αα

=+

22sin cos2 22cos 2

α α

α

⋅= tan 2α=

Alternatively : Substituting 1,n = in given expression we get, sin 2 0cos 2 2cos 1αα α

−+ + 2

2sin cos2 22cos 2α α

α

⋅=

22sin cos sin tan2cos 2cos cos 1 2α α α αα α α

= = =+ +

83. Ans. (d), ( )

( )sin 70 cos 90 50sin 70 cos 40cos 70 sin 40 cos 70 sin 90 50° − °− °° − °

=°+ ° °+ ° − °

sin 70 sin 50cos 70 cos50°− °

=° + °

70 50 70 502cos sin2 270 50 70 502cos cos2 2° + ° ° − ° ⋅

=° + ° ° − ° ⋅

2cos60 sin10 sin10 tan102cos60 cos10 cos10° ° °= = = °

° ° °

84. Ans. (a), ∵ cos cossin sinA B

A B

+−

2cos cos2 22cos sin2 2A B A B

A B A B

+ −

=+ −

cot 2A B−=

and 2sin cossin sin 2 2cos cos 2sin sin2 2A B A B

A B

A B A BA B

+ −+

=+ −− −

cot 2A B−= −

Therefore cos cos sin sinsin sin cos cosn n

A B A B

A B A B

+ + + − − ( )cot 1 cot2 2nn nA B A B− − = + −

But given, n is odd, ∴ ( )1 1n− = − . Therefore, the value of the expression is 0.

85. Ans. (c), Given, cot cot 2α β = …(1) Now ( )

( )cos cos cos sin sincos cos cos sin sinα β α β α β

α β α β α β+ −

=− +



( )( )

cos 1 cot cotcos 1 cot cotα β α βα β α β+ −

⇒ =− +

1 2 11 2 3− −= =

+ [Dividing Nr. and Dr. by cos cosα β⋅ ]

86. Ans. (b), cos 0 cos1 ... cos89 cos90 cos91 ... cos178 cos179° + ° + + °+ ° + ° + + °+ ° ( ) ( ) ( )1 cos1 cos 2 ... cos89 0 cos 180 89 .. cos 180 2 cos 180 1= + °+ °+ + °+ + °− ° + + °− ° + °− ° 1 cos1 cos 2 ... cos89 cos89 cos88 ... cos 2 cos1 1= + °+ ° + + °− °− ° − ° − ° = 87. Ans. (b), Given 2 2 2cos cos 1 sinA B C= + + − ( ) ( )2cos cos cos 1A B C B C= + + − + ( ) ( )21 sin cos 90 cos 1A A B C= − + °− − + ( )22 sin sin cosA A B C= − + − ( )( )2 sin sin cosA A B C= − − − ( ) ( )2 sin cos cosA B C B C= − + − − 2 sin 2sin sinA B C= + ⋅ 2 2sin sin sinA B C= + Alternatively : By setting 30A B C= = = ° , we get 2 2 2cos cos cosA B C+ + 22 3 93cos 30 3 2 4

= ° = =

Option (b) gives 2 2sin 30 sin 30 sin 30+ ° ° ° 1 1 1 1 92 2 22 2 2 4 4 = + = + =

88. Ans. (c), ( ) ( ){ }tan 2 tanA A B A B= + + −( ) ( )( ) ( )

tan tan .1 tan tan 1A B A B p q

A B A B pq

+ + − += =

− + − −

89. Ans. (c), cos10 sin10 1 tan10cos10 sin10 1 tan10° + ° + °=

° − ° − ° [Dividing Nr. & Dr. by cos10° ]

tan 45 tan101 tan 45 tan10°+ °=

− ° ° [Using tan 45 1° = ]

( )tan 45 10= °+ ° ( )tan tan tan1 tan tanA BA B

A B

+ = + − ∵

tan 55= ° 90. Ans. (c), 312sin 40 16sin 40° − ° ( )34 3sin 40 4sin 40= °− ° ( ) ( )4sin 3 40 4sin120 4sin 180 60 4sin 60= × ° = ° = °− ° = ° 34 2 3.2= × = 91. Ans. (c), ( ) tan tantan 1 tan tanA B

A BA B

++ =

− 1 12 31 11 2 3

+=

− ×

1 1tan , tan2 3A B = = ∵

5 5 / 66 11 5 / 61 6= = = −

( )tan 1 tan 4Cπ

π⇒ − = = [ ]A B C A B Cπ π+ + = ⇒ + = −∵ 4C

ππ⇒ − = 34 4C

π ππ⇒ = − =



92. Ans. (b), ∵ 3 tan 4 0A− = 4tan 3A⇒ = 2422 tan 8 9 243sin 2 .161 tan 3 25 251 9

AA

A

×∴ = = = × =

+ +

Also, 2 16 5sec 1 tan 1 9 3A A= − + = − + = − rdsec is ve since lies in III quadrantA A − 3cos 5A∴ = − and so 2 4sin 1 cos 5A A

−= − − =

Now, 5sin 2 3sin 4cosA A A+ + 24 4 35 3 425 5 5− − = × + × + ×

24 12 12 05 5 5= − − = 93. Ans. (c), 8cos 17θ = 2sin 1 cosθ θ⇒ = − 64 151 .289 17= − = stNote : sin is ve since lies in 1 quadrantθ θ + Now, ( ) ( ) ( )cos 30 cos 45 cos 120θ θ θ°+ + °− + °− cos30 cos sin 30 sin cos 45 cos sin 45 sin cos120 cos sin120 sinθ θ θ θ θ θ= ° − ° + ° + ° + ° + ° 3 1 1 1 1 3cos sin cos sin cos sin2 2 2 22 2θ θ θ θ θ θ= − + + − + ( ) ( ) ( )3 1 1sin cos sin cos sin cos2 2 2θ θ θ θ θ θ= + − + + + ( ) 3 1 1sin cos 2 2θ θ

−= + +

15 8 3 1 117 17 2 2 − = + + 23 3 1 1 .17 2 2 −= +

94. Ans. (c), 2 2sin 17.5 sin 72.5°+ ° ( )2 2sin 17.5 sin 90 17.5= °+ °− ° 2 2 2 2sin 17.5 cos 17.5 1 1 tan 45= °+ ° = = = ° [ ]tan 45 1° =∵ 95. Ans. (d), 3 34sin cos 4cos sinA A A A− ( )2 24sin cos cos sinA A A A= − ( )2 2sin cos .cos 2 2sin 2 .cos 2 sin 4 .A A A A A A= = = 96. Ans. (c), 3 3cos cos3 sin sin 3cos sinθ θ θ θ

θ θ− +

+ ( ) ( )3 3 3 3cos 4cos 3cos sin 3sin 4sincos sinθ θ θ θ θ θ

θ θ

− − + −= +

3 33cos 3cos 3sin 3sincos sinθ θ θ θθ θ

− + −= + ( ) ( )2 2 2 23 1 cos 3 1 sin 3sin 3cosθ θ θ θ= − + − = +

( )2 23 sin cos 3 1 3.θ θ= + = × = 97. Ans. (d), ( )

( )

22 22 22

1 2 tan 11 tan 2 tancos 2 1 tan 2 tan 21 2 tan 1βα βα

α ββ

− +− −= = =

+ ++ +

( )2 2 2222 tan tan sin .sec2 tan 1β β

βββ

− −= = = −

+

98. Ans. (a) 2 2A CA C B B

++ = ⇒ =

Now, 2sin sincos cos 2 2 tansin sin 2cos sin2 2A C A C

C A A C

A C A CA C

+ − − + = = + −−



tan B= [Using (i)] 99. Ans. (c), Given, 1cos 7P = 2 1sin 1 cos 1 49P P⇒ = − = −

48 4 349 7= = [Since Q is acute, so sin P is positive] and 13cos 14Q = 2 169sin 1 cos 1 196Q Q⇒ = − = −

27 3 3196 14= = [Since Q is acute, so sinQ is positive] Now, ( )sin sin cos cos sinP Q P Q P Q− = −

4 3 13 1 3 3 52 3 3 3. .7 14 7 14 98−= − =49 3 3 sin 6098 2= = = °

60 .P Q⇒ − = ° 100. Ans. (b), sin120 cos150 cos 240 sin 330° ° − ° ° ( ) ( ) ( ) ( )sin 180 60 cos 180 30 cos 180 60 sin 360 30= °− ° ° − ° − °+ ° ° − ° ( ) ( ) ( )( )sin 60 . cos30 cos60 sin 30= ° − ° − − ° − ° 3 3 1 1. .2 2 2 2 = − − − −

3 1 4 14 4 4= − − = − = − 101. Ans. (c), 4 4cos sin24 24π π −

2 2 2 2cos sin . cos sin24 24 24 24π π π π = − +

3 1cos 2 .1 cos cos1524 12 2 2π π + = = = ° =

102. Ans. (c), sin 300 tan 330 sec 420tan135 sin 210 sec315° ° °

° ° ° ( ) ( ) ( )

( ) ( ) ( )sin 360 60 tan 360 30 sec 360 60tan 180 45 sin 180 30 sec 360 45° − ° ° − ° ° + °

=° − ° ° + ° °− °

{ } { }

{ } { }sin 60 . tan 30 sec 60tan 45 . sin 30 .sec 45− ° − ° °

=− ° − ° °

3 1. .2sin 60 .tan 30 .sec60 2 3 21tan 45 sin 30 .sec 45 1. . 22° ° °

= = =° ° °

103. Ans. (a) 3 5 71 cos 1 cos 1 cos 1 cos8 8 8 8π π π π + + + +

3 31 cos 1 cos 1 cos 1 cos8 8 8 8π π π π = + + − −

5 3 3cos cos cos and 8 8 87cos cos cos8 8 8π π π

π

π π ππ

= − = −

= − = −

∵ 2 2 31 cos 1 cos8 8π π = − −

2 2 3sin .sin8 8π π

= 2 2sin .cos8 8π π

= 3 3sin cos cos8 2 8 8π π π π = − =

∵ 2 21 12sin cos sin [ 2sin cos sin 2 ]4 8 8 4 4 A A A

π π π = = =

∵ 21 1 1.4 82

= =

. 104. Ans. (b), 90x y+ = ° 90y x⇒ = °− ( ) 1sin . sin sin .sin 90 sin .cos sin 2 .2x y x x x x x= °− = =∵



Now, 1 sin 2 1x− ≤ ≤ 1 1 1sin 22 2 2x⇒ − ≤ ≤ 1 1sin .sin .2 2x y⇒ − ≤ ≤ ∴ Maximum value of 1sin .sin is 2x y . 105. Ans. (c), Given, cot cot nα β− = …(i) 1 1 cot cottan tan cot cot cot cotβ α

α βα β α β

−∴ − = − = cot cotnm

α β−

⇒ = [Using (i)] cot cot n

mα β

−⇒ = …(ii)

Now, ( )1cot cot 1 1 1cot cot cot

n

m

n m n

α βα β

β α

−++

− = = = −− −

[Using (i) and (ii)] 106. Ans. (c), 1sec tan 1m

mθ θ

+− =

− …(i)

By taking inverse on both sides we get, 1sec tan 1m

mθ θ

−+ =

+ …(ii)

Adding (i) and (ii) we get, ( )( )

( )( )

( )( )

22

2 11 12sec 1 1 1mm m

m m mθ

++ −= + =

− + −

22 1sec 1m

mθ

+⇒ =

− 2

2 1cos .1m

mθ

−⇒ =

+

107. Ans. (b), ( ) ( )tan 2 tanα β α α β+ = + + tan tan tan tan tan tan1 tan tan tan tan tan tanα α β α α βα α α β β α+ + −

=− − −

( ) tan tan tan tan tan tantan 1 tan tan tan tan tan tanA B C A B C

A B CA B B C C A

+ + − + + = − − − ∵

222 tan tan tan . tan1 tan 2 tan tanα β α βα α β

+ −=

− −

22

1 1 1 12. .3 7 3 71 1 11 2. .3 3 7 + − = − −

502 1 1 633 7 63 1 tan1 2 50 41 9 21 63π

+ −

= = = = − −

( )2 4πα β⇒ + = 108. Ans. (d), 212cot 31cosec 32 0θ θ− + = 2 212cos 31sin 32sin 0θ θ θ⇒ − + = ( )2 212 1 sin 31sin 32sin 0θ θ θ⇒ − − + = 220sin 31sin 12 0θ θ⇒ − + = 31 961 960 31 1 3 4sin or40 40 4 5θ

± − ±⇒ = = =

109. Ans. (d), Given, sec tan sec tanθ θ θ θ= + − = For checking choice (a) Given cosec sec tanθ θ θ= + − ∴ It is wrong For checking choice (b) Given ( )cosec sec tanθ θ θ− − cosec sec tanθ θ θ= − + ∴ It is wrong For checking choice (c) Given sec sec tanθ θ θ= + − ∴ It is wrong For checking choice (d) Given ( )sec sec tan tanθ θ θ θ= − − = , Hence choice (d) is correct



110. Ans. (a), 22 .sin 12x xα+ = 2 1sin 2 2x xα −

⇒ = 1 cos 12 2x xα− − ⇒ =

21 cos 2 2sinA A − = ∵ 1cos

xα⇒ = x⇒ must be positive

Now 22 21 1sin 1 cos 1 x

x xα α

−= − = − = and so, 2sintan 1.cos x

αα

α= = −

[Note sin , cosα α and tanα are all positive since α is acute] 111. Ans. (c), 2sin sin 2 sin 2sin cos1 cos cos 2 cos 2cosθ θ θ θ θ

θ θ θ θ+ +

=+ + +

21 cos 2 2cosθ θ + = ∵ ( )

( )sin 1 2cos sin tan .cos 1 2cos cosθ θ θ

θθ θ θ

+= = =

+

112. Ans. (a) , 2 21 2cos 12cot sin sin sinxx

x x x+ = + 22sin cos 1sinx x

x

+=

( )22 22 2sin cos2sin cos sin cossin sinx xx x x x

x x

++ += = sin cos sin cossin sinx x x x

x x

+ − −= =

( )

3When , we have, sin 0 but cos 0 and cos is greater than sin in4magnitude i.e. cos sin and so sin cos is ve sin cos sin cosx x x x x

x x x x x x x x

ππ < < ≥ <

> + − ∴ + = − +

∵ 113. Ans. (a), cos1 cos 2 cos3 ..... cos180°+ ° + ° + + ° ( ) ( ) ( ) ( ){ }cos1 cos179 cos 2 cos178 cos3 cos177 ... cos89 cos91= °+ ° + °+ ° + °+ ° + + °+ ° cos90 cos180+ °+ ° ( ) ( ) ( ) ( ) ( )cos1 cos1 cos 2 cos 2 cos3 cos3 ... cos89 cos89 0 1= °− ° + °− ° + °− ° + + °− ° + + − ( )cos 180 cosθ θ° − = − ∵ ( ) ( )0 0 0 .... 0 0 1 1.= + + + + + + − = − 114. Ans. (c), ∵ 5cos 3 0C + = 3cos 5C⇒ = − Since C is an angle of a ABC∆ and cosC is ,ve− we have .2 C

ππ< <

Now, 2 9 4sin 1 cos 1 25 5C C= − = − = sin is ve since 2C Cπ

π + < < ∵

sin 4 / 5 4tan cos 3/ 5 3CC

C∴ = = = −

−, We have 4 4 8sin tan 5 3 15C C

− + = + − =

(Sum of the roots) and 4 4 16sin .tan .5 3 15C C

= − = −

(Product of the roots) ∴ Required quadratic equation is 2 8 16 015 15x x

− − + − =

215 8 16 0x x⇒ + − = . 115. Ans. (c), ( )sin 1α β+ = 2πα β⇒ + = …(i) ( ) 1sin 2α β− = 6πα β⇒ − = …(ii) Solving (i) and (ii) we get ,3 6π π

α β= = ( ) ( )tan 2 .tan 2α β α β∴ + +



1/2p = 1p =−+ +

( )( )2 1 1p p− −

2tan .tan3 3 3 6π π π π = + +

2 5tan . tan tan .tan3 6 3 6π π π ππ π = = − −

( ) 1tan . tan 3 13 6 3π π = − − = − − =

116. Ans. (d), For checking choice (a) ( )2 2 2cot tan cot tan 2θ θ θ θ+ = + + ( ) ( )2 2 2 2cosec 1 sec 1 2 sec cosecθ θ θ θ= − + − + = + , Hence, option (a) is correct. For checking choice (b) ( )

22 2 22 cos sin cos sincot tan sin cos sin cosθ θ θ θθ θ

θ θ θ θ + + = + =

2 2 21 sec coseccos sin θ θθ θ

= = ⋅

, Hence, option (b) is correct. For checking choice (c) ( )

22 2 22 cos sin cos sincot tan sin cos sin cosθ θ θ θθ θ

θ θ θ θ − − = − =

( )

22 2cos 2 2cot 2 4cot 21 sin 22

θθ θ

θ

= = =

, Hence, option (c) is correct, Hence, option (d) is correct 117. Ans. (a), 22sin 3sin 1 0x x− + ≥ 22sin 2sin sin 1 0x x x⇒ − − + ≥ ( ) ( )2sin sin 1 1 sin 1 0x x x⇒ − − − ≥ ( )( )2sin 1 sin 1 0x x⇒ − − ≥ ( )( )2 1 1 0p p⇒ − − ≥ where sinp x= 1 or 12p p⇒ ≤ ≥ 1sin or sin 12x x⇒ ≤ ≥ 1sin 2x⇒ ≤ sin can not be greater than 1 and since 0, , so sin 12x x x

π ∈ ≠ ∵

0 6xπ

⇒ ≤ ≤ 118. Ans. (b), 2 2 2

2 2 21 sin sin sincos 1 cos cos 04sin 4 4sin 4 1 4sin 4θ θ θθ θ θθ θ θ

+

+ =

+

[ ]1 1 3 2 2 3,C C C C C C→ − → −

221 0 sin0 1 cos 01 1 1 4sin 4θθθ

⇒ =

− − +

( ) ( ){ } ( )2 21 4sin 4 cos 0 sin 0 1 0θ θ θ ⇒ + − − − + + = ( ) 2 21 4sin 4 cos sin 0θ θ θ⇒ + + + = 1 4sin 4 1 0θ⇒ + + = 1sin 4 2θ⇒ = − 119. Ans. (c), 2 2sin sin8 2 8 2A Aπ π + − −

sin sin8 2 8 2 8 2 8 2A A A Aπ π π π = + + − ⋅ + − +

sin sin4 Aπ

= ( ) ( ) 2 2sin sin sin sinA B A B A B + − = − ∵ 1 sin .2 A=



120. Ans. (b), 22 21 tan 2 tancos 2 sin 2 1 tan 1 tanx x

a x b x a bx x

− + = + + +

( ) ( )2

2 2222

1 2.1 tan 2 tan1 tan 1b b

a ba x b x a a

bx

a

− + − + = =+

+

2 2 22 2 2

2 22 22

2.

a b b

a a a b aa

a a ba b

a

−+ + = = ⋅ =

+ +

121. Ans. (b), tan cot mα α+ = ( )3 3tan cot mα α⇒ + = ( )3 3 3tan cot 3 tan cot tan cot mα α α α α α⇒ + + + = 3 3 3tan cot 3.1.m mα α⇒ + + = [ ]tan cot 1α α =∵ ( )3 3 3 2tan cot 3 3 .m m m mα α⇒ + = − = − 122. Ans. (c), 2 2cos sin6 6π π

θ θ + − −

cos cos6 6 6 6π π π π

θ θ θ θ = + + − + − +

( ) ( )2 2cos sin cos cosA B A B A B − = + − ∵ 1cos . cos 2 cos 2 .3 2π

θ θ= = 123. Ans. (c), We have, sin 2 sin 2 sin 2A B C+ + 2 2 2 22sin .cos sin 22 2A B A B

C+ − = +

( ) ( )2sin cos 2sin cosA B A B C C= + − + ( )2sin cos 2sin cosC A B C C= − + ( ) ( )sin sin sinA B C Cπ+ = − = ∵ ( ) ( )2sin cos cosC A B A B= − − + ( ){ } ( )cos cos cosC A B A Bπ = − + = − + ∵ [ ]2sin 2sin sin 4sin sin sin .C A B A B C= = Also, cos cos cos 1A B C+ + − 22cos cos 1 2sin 12 2 2A B A B C+ − = + − −

22sin cos 2sin2 2 2C A B C− = −

cos cos sin2 2 2 2A B C Cπ + = − = ∵

2sin cos cos2 2 2C A B A B − + = −

sin sin cos2 2 2 2C A B A Bπ + + = − =

∵ 4sin sin sin 4sin sin sin2 2 2 2 2 2C A B A B C

= = sin 2 sin 2 sin 2cos cos cos 1A B C

A B C

+ +∴

+ + − 4sin sin sin 8cos cos cos2 2 24sin sin sin2 2 2

A B C A B C

A B C= =

(�) BY R . K . MAL IK ’S NEWTON CLASSES

21


124. Ans. (c), 2 2 2 22sin .cos sin 22 2A B A BC

+ − = +

sin 2 sin 2 sin 2A B C+ + ( ) ( )2sin cos 2sin cosA B A B C C= + − + ( ) ( )2sin cos 2sin cosC A B C Cπ= − − + ( ){ }2sin cos cosC A B C= − + ( )sin sinC Cπ − = ∵ ( ) ( )( ){ }2sin cos cosC A B A Bπ= − + − + ( ) ( ){ }2sin cos cosC A B A B= − − + { }2sin 2sin sin 4sin sin sinC A B A B C= = . 125. Ans. (c), 2cos cos3 cos5θ θ θ− − ( ) 3 5 3 52cos cos3 cos5 2cos 2cos cos2 2θ θ θ θ

θ θ θ θ+ − = − + = − ⋅

{ }2cos 2cos 4 cosθ θ θ= − ( ) ( )22cos 1 cos 4 2cos 2sin 2θ θ θ θ= − = 21 cos 2 2sinA A − = ∵ ( )22 2 34cos sin 2 4cos 2sin cos 16sin cos .θ θ θ θ θ θ θ= = = 126. Ans. (a), We have, 2cos 2cos3 cosx x y+ = …(i) 2sin 2sin 3 sinx x y+ = …(ii) Squaring and adding (i) and (ii) we get ( ) ( )2 2 2 2 2 24cos 4cos 3 8cos cos3 4sin 4sin 3 8sin sin 3 cos sinx x x x x x x x y y+ + + + + = + ( ) ( ) ( )2 2 2 24 cos sin 4 cos 3 sin 3 8 cos cos3 sin sin 3 1x x x x x x x x⇒ + + + + + = ( )4 4 8cos 3 1x x⇒ + + − = ( )cos cos sin sin cosA B A B A B+ = − ∵ 8 8cos 2 1x⇒ + = 7cos 2 8x

−⇒ =

127. Ans. (a),∵ 2 1 1cos 1 sin 12 2 2 2x x

x x

θ θ − += − = − =

Note : 0 0 cos is ve and tan is also ve2 2 4 2 2π θ π θ θθ < < ⇒ < < ∴ + +

∴ 2

1 12. .2sin cossin 2 22 2tan 1cos 1 2sin 1 2.2 2x x

x x

x

x

θ θθ

θθθ

− +

= = =− − −

2 22 1 12 2 2xx

x x

−= = −

− +

128. Ans. (d), sin 3 sin 5 sin 7 sin 9cos3 cos5 cos 7 cos9θ θ θ θθ θ θ θ

+ + ++ + +

( ) ( )( ) ( )

sin 3 sin 9 sin 5 sin 7cos3 cos9 cos5 cos7θ θ θ θθ θ θ θ

+ + +=

+ + +

( )( )

2sin 6 cos3 cos2sin 6 cos3 2sin 6 cos2cos6 cos3 2cos 6 cos 2cos 6 cos3 cosθ θ θθ θ θ θθ θ θ θ θ θ θ

++= =

+ +tan 6 .θ=

129. Ans. (d), Given, ( )cos sinsin cosA

α αα

α α

= − , ∴ ( ) ( )

cos sin cos sin. sin cos sin cosA Aα α β β

α βα α β β

= − −

cos cos sin sin cos sin sin cossin cos cos sin sin sin cos cosα β α β α β α β

α β α β α β α β− +

= − − − + ( ) ( )

( ) ( )( )

cos sin .sin cos Aα β α β

α βα β α β+ +

= = + − + +

130. Ans. (d), cos15 sin15 cos15 cos 75 2sin 45 sin 30° − ° = °− ° = ° ° cos cos 2sin sin2 2C D D C

C D + − − =

∵



1 1 12 . 22 2= × = 131. Ans. (c), 1 1sin cos 2 sin cos2 2x x x x

+ = +

2. sin cos cos sin4 4x x

π π= +

2 sin 4xπ = +

, As 1 sin 14x

π − ≤ + ≤

[ ]1 sin 1θ− ≤ ≤∵ 0 sin 14x

π ⇒ ≤ + ≤

0 2 sin 24xπ ⇒ ≤ + ≤

0 sin cos 2.x x∴ ≤ + ≤

132. Ans. (d), ( ) ( )( )

323tan / 3 tan / 3 tan 3 tan1 3tan / 3 3A A A

AA

− = = − 3

23tan tantan 3 1 3tanθ θθ

θ −

= − ∵

133. Ans. (d), 1 0 2cos 1 01 1 1 2cos 16 1 6 1 2cosC

C

C

θ

θθ

∆ = = ( ) ( )22cos 4cos 1 1 2cos 6θ θ θ= − − − 3 38cos 2cos 2cos 6 8cos 4cos 6θ θ θ θ θ= − − + = − + (b) and (c) are wrong choices can be checked easily by supposing 2πθ = Hence, given expression become 6 but sin 4sin 4θθ becomes 0 and 22sin 2sin θ

θ also becomes 0

134. Ans. (d), cos 20 . cos 40 . cos60 . cos80° ° ° ° 1 cos 20 . cos 40 . cos802= ° ° ° ( ) ( )1 .cos 20 . cos 60 20 . cos 60 202= ° °− ° ° + ° ( )1 1. cos 3 202 4= × ° ( ) ( ) 1Using the formula cos .cos 60 cos 60 cos34θ θ θ θ °− ° + =

1 1 1 1cos 60 .8 8 2 16= ° = × = 135. Ans. (d), Since , ,α β γ are in A.P., so we have 2α γ β+ = 2α γ

β+

⇒ = …(i) Now, 2cos sinsin sin 2 2cos cos 2sin sin2 2

α γ α γα γ

α γ α γγ α

+ − − =

+ −−

cos cos2 sinsin 2α γ

βα γ β

+ = =

+

[Using(i)] cot .β= 136. Ans. (a), ( )cos52 cos68 cos172°+ ° + ° ( ) ( )2cos 60 cos8 cos 180 8= ° ° + °− ° 12 cos8 cos8 cos8 cos8 0.2 = × × ° − ° = ° − ° =

137. Ans. (c), 2 2 2cos cos sinA B C+ − 1 cos 2 1 cos 2 cos 2 12 2 2A B C+ + −= + +

( )1 1 cos 2 cos 2 cos 22 2 A B C= + + + ( ) ( )( )21 1 2cos cos 2cos 12 2 A B A B C= + + − + − ( ) ( ) 2cos cos cosA B A B C= + − + ( ) ( ) 2cos 2 cos cosC A B Cπ= − − +


23


( ){ }cos cos cosC A B C= − + ( )cos 2 cosC Cπ − = ∵ ( ) ( ){ }cos cos cosC A B A B= − + + ( ){ } ( )cos cos 2 cosC A B A Bπ = − + = + ∵ { }cos 2cos cosC A B= 2cos cos cos .A B C= 138. Ans. (b), Given, ( )2sec tanx a θ θ= + ( )2sec tan x

aθ θ⇒ + = …(i)

Now, ( ) ( )( )

( )44 44sec tansec tan . sec tansec tanb

y bθ θ

θ θ θ θθ θ

−= − = +

+

( )( ) ( )

42 24 4 2sec tansec tan sec tanb b b

x

a

θ θ

θ θ θ θ

−= = =

+ +

[Using(i)] 2 2x y a b⇒ = 4 2 4 2.x y a b⇒ = 139. Ans. (d) 2 2(1 tan tan ) (tan tan )A B A B+ + − 2 2 2 21 tan tan 2 tan tan tan tan 2 tan tanA B A B A B A B= + + + + − 2 2 2 21 tan tan tan tanA A B B= + + + 2 2 2 2 2(1 tan ) tan (1 tan ) (1 tan ) (1 tan )A B A A B= + + + = + + 2 2sec .sec .A B= 140. Ans. (c), Let 23sin cos 2 3sin 1 2sinP θ θ θ θ= + = + − 2 32 sin sin 12θ θ = − − +

2 3 9 92 sin sin 12 16 8θ θ = − − + + +

23 172 sin .4 8θ = − − +

Clearly, P has the maximum value when 23sin 4θ −

has the

Minimum value i.e., 0. 23sin 04θ − ≥

∵ ∴ The maximum value of P is 17 .8

141. Ans. (a), 2 22sin sin15 30π π−

2 2sin sin sin sin15 30 15 30 6 10π π π π π π = + − =

( ) ( )2 2sin sin sin sinA B A B A B − = + − ∵ 1 5 1 5 1.2 4 8− −

= = 5 1sin sin1810 4π −= ° =

∵

142. Ans. (d), ( )3 2 2 2 2

13 5cos 2 1 cos cos cos12 12 12 12k

kπ π π π

=

− = + +

∑ 2 2 2cos cos sin12 4 2 12π π π π = + + −

22 21 1 3cos sin 1 .12 12 2 22π π = + + = + =

143. Ans. (d), ( ) ( )24 4 2 2 2 2sin cos sin cos 2sin cosf x x x x x x x= + = + − ( )211 sin 2 .2 x= − Clearly, ( )f x has a minimum value when ( )2sin 2x has the maximum value.



i.e., when sin 2x has a maximum value 0 0 2 sin 2 is ve2x x xπ

π ≤ ≤ ⇒ ≤ ≤ ⇒ + ∵

i.e., when sin 2 1x = i.e., when 2 i.e.2xπ

= when .4xπ

= ∴ Minimum value of ( )f x 4 44 4 1 1sin cos4 4 4 2 2f

π π π = = + = +

1 1 14 4 2= + = Alternatively : ( ) ( )24 4 2 2 2 2sin cos sin cos 2sin cosf x x x x x x x= + = + − ( )211 sin 2 .2 x= − Clearly, ( )f x has a minimum value when ( )21 sin 22 x has a maximum value. We have 1 sin 2 1x− ≤ ≤ ( )20 sin 2 1x⇒ ≤ ≤ ( )21 10 sin 22 2x⇒ ≤ ≤ ∴ Maximum value of ( )21 sin 22 x is 12 and so, maximum value of ( ) 1 11 .2 2f x = − = 144. Ans. (d), 2 2 2sin sin sinA B C+ + 1 cos 2 1 cos 2 1 cos 22 2 2A B C− − − = + +

( )3 1 cos 2 cos 2 cos 22 2 A B C= − + + ( ) ( ) ( ){ }23 1 2cos 1 2cos cos2 2 A B C B C= − − + + − ( ) ( ){ }23 1 1 2cos 2cos cos2 2 A A B Cπ= − − + + − − [ ]A B C B C Aπ π+ + = ⇒ + = −∵ ( ){ }212 2cos 2cos cos2 A A B C= − − − ( ) ( ){ }2 cos cos cosA B C B C= + ⋅ + + − ( ){ } ( )cos cos cosA B C B Cπ = − + = − + ∵ 2 2cos cos cos .A B C= + 145. Ans. (d), Since ,2π α π< < so sinα is ve+ and cosα is ve− Since 3 ,2ππ β< < so sin β is ve− and cos β is ve− Now, 2 225 64 8cos 1 sin 1 289 289 17α α= − − = − − = − = − [ ]cos is veα −∵ 2 144 169 13sec 1 tan 1 25 25 5β β= − + = − + = − = − [ ]cos is ve and also sec is veβ β− −∵ 1 5cos sec 13β

β∴ = = − and so, 2 25 144 12sin 1 cos 1 169 169 13β β= − − = − − = − = − [ ]sin is veβ −∵

Now, ( )sin sin cos cos sinβ α β α β α− = − 12 8 5 15 96 75 171. .13 17 13 17 221 221+ = − − − − = =

146. Ans. (a), 2 2 1cos 2 2cos 2cos 1 2cos 2 cos cos 2θ θ θ θ θ θ + = − + = + −

2 21 3 1 32 cos 2 cos2 4 2 2θ θ = + − = + −

Now, we have 21cos 02 Rθ θ + ≥ ∀ ∈


25


Also, maximum value of 2 21 1 9cos is 1 i.e.,2 2 4θ + +

[Obtained when cos 1θ = ] 21 90 cos 2 4θ ∴ ≤ + ≤

21 90 2 cos 2 2θ ⇒ ≤ + ≤

23 1 3 9 30 2 cos2 2 2 2 2θ ⇒ − ≤ + − ≤ −

3 cos 2 2cos 32 θ θ⇒ − ≤ + ≤ 147. Ans. (d), ( ) ( )2 2 2cos cos 120 cos 120α α α+ − ° + + ° { } { }2 22cos cos cos120 sin sin120 cos cos 1̀20 sin sin120α α α α α= + °+ ° + ° − ° 2 22 1 3 1 3cos cos sin cos sin2 2 2 2α α α α α

= + − + + − −

2 2 21 3 3cos cos sin sin cos4 4 2α α α α α= + + − 2 21 3 3cos sin sin cos4 4 2α α α α+ + + ( )2 2 2 23 3 3 3 3cos sin cos sin 1 .2 2 2 2 2α α α α= + = + = × = 148. Ans. (b), We know that tan tan tan tan tan tan 12 2 2 2 2 2α β β γ γ α

⋅ + ⋅ + ⋅ = if 180α β γ+ + = ° Hence, here tan tan tan tan tan tan 1A B B C C A+ + = [∵ 90A B C+ + = ° ] tan tan tan tan tan tan 1A B B C C A⇒ + + = 149. Ans. (a), ( ) ( )6 6 4 42 sin cos 3 sin cos 1θ θ θ θ+ − + + ( ) ( ){ }32 2 2 2 2 22 sin cos 3sin cos sin cosθ θ θ θ θ θ= + − + ( ){ }22 2 2 23 sin cos 2sin cos 1θ θ θ θ− + − + ( ) ( ) ( ) ( ) ( )3 23 3 2 2Using 3 and 2a b a b ab a b a b a b ab + = + − + + = + − { } { }2 2 2 22 1 3sin cos 3 1 2sin cos 1 0θ θ θ θ= − − − + = 2 2 2 22 6sin cos 3 6sin cos 1θ θ θ θ= − ⋅ − + + 150. Ans. (b), ( ) { }

sin 2 3sin 5 1 sin 2 cos3 cos 2 sinsin sin sinθ θθθ θ θ θ

θ θ θ+

= = + 3 ( ) ( )( ){ }3 2 31 2sin cos 4cos 3cos 2cos 1 3sin 4sinsin θ θ θ θ θ θ θ

θ= − + − −

( ) ( ) ( )3 2 22cos 4cos 3cos 2cos 1 . 3 4sinθ θ θ θ θ= − + − − ( ) ( ){ }4 2 2 28cos 6cos 2cos 1 . 3 4 1 cosθ θ θ θ= − + − − − ( )( )4 2 2 28cos 6cos 2cos 1 4cos 1θ θ θ θ= − + − − 4 2 4 28cos 6cos 8cos 6cos 1θ θ θ θ= − + − + 4 216cos 12cos 1.θ θ= − + 151. Ans. (d), 21 cos cos .... 2 2α α+ + + = − 1 2 21 cosα⇒ = −

− 21 cos cos ....is an infinite G.P. with common ratio1cos 1 and first term 1 and so 1 1 cosa

r a Sr

α α

αα∞

+ + + = < = = = − −

∵ 11 cos 2 2α⇒ − =

− 1 2 2 2 2 2 2 1. 14 2 22 2 2 2 2+ + += = = = +

−− +

1 3cos cos cos cos4 4 42 π π πα π ⇒ = − = − = − =

3 .4πα⇒ =



152. Ans. (d), 2 21 1cos 7 cos 372 2 ° − °

2 2 2 21 1 1 11 sin 7 1 sin 37 sin 37 sin 72 2 2 2 = − ° − + ° = ° − °

( ) ( )sin 45 sin 30= ° ° ( ) ( )2 2sin sin sin sinA B A B A B − = + − ∵ 1 1 1. .22 2 2= = 153. Ans. (b), 3 5 7 9tan tan tan tan tan20 20 20 20 20π π π π π

3 3tan tan tan tan tan20 20 4 2 20 2 20π π π π π π π = − −

3 3tan tan .1.cot .cot 1.20 20 20 20π π π π = =

154. Ans. (b), 2 2 2 5cos cos cos12 4 12π π π+ + 2 2 2 5sin cos cos2 12 4 12π π π π = − + +

sin cos2π θ θ

− = ∵

22 25 1 5 1sin cos 112 12 22π π = + + = +

2 2sin cos 1A A + = ∵

3 .2= 155. Ans. (a), sin sin sin 2sin cos sin2 2A B A B

A B C C+ − + + = +

2sin cos 2sin cos2 2 2 2C A B C C− = − +

sin sin sin2 2 2A B C C + = − = − ∵

2sin cos cos2 2 2C C A B − = −

2sin cos cos2 2 2C A B A B + − = − cos cos cos2 2 2C A B A B + + = − =

∵

2sin . 2sin sin2 2 2C A B = − 4sin sin sin .2 2 2A B C= −

156. Ans. (b) cos35 cos85 cos155° + ° + ° 2cos 60 cos 25 cos(180 25 )= ° ° + °− °

12. .cos 25 cos 25 cos 25 cos 25 02= °− ° = ° − ° = 157. Ans. (b), ( )sin 1A B C+ + = 90A B C⇒ + + = ° …(i) ( ) 1tan 3A B− = 30A B⇒ − = ° …(ii) ( )sec 2A C+ = ( ) 1cos 2A C⇒ + = 60A C⇒ + = ° …(iii) Subtracting (iii) from (i) we get 30 .B = ° Substituting 30B = ° in (ii) we get 60 .A = ° Substituting 60A = ° in (iii) we get 0 .C = ° 158. Ans. (c) (a) When ,A B C π+ + = we have sin sin sin 2sin cos 2sin cos2 2 2 2A B A B C C

A B C+ − + + = +

2cos cos 2sin cos2 2 2 2C A B C C− = +

sin sin cos2 2 2 2A B C Cπ + = − = ∵


27


2cos cos cos2 2 2C A B A B − + = +

sin sin cos2 2 2 2C A B A Bπ + + = − =

∵ 2cos 2cos cos 4cos cos cos2 2 2 2 2 2C A B A B C = =

∴ Statement (a) is correct.

(b) When 2 ,A B C π+ + = we have, sin sin sin 2sin cos 2sin cos2 2 2 2A B A B C CA B C

+ − + + = +

2sin cos 2sin cos2 2 2 2C A B C C− = +

sin sin sin2 2 2A B C C

π + = − =

∵

2sin cos cos2 2 2C A B C − = +

2sin cos cos2 2 2C A B A B − + = −

cos cos cos2 2 2C A B A B

π + + = − = −

∵

2sin 2sin sin 4sin sin sin2 2 2 2 2 2C A B A B C = =

∴ Statement (b) is correct. (c) When 0,A B C+ + = we have sin sin sin 2sin cos 2sin cos2 2 2 2A B A B C C

A B C+ − + + = +

2sin cos 2sin cos2 2 2 2C A B C C− = − +

sin sin sin2 2 2A B C C + = − = − ∵

2sin cos cos2 2 2C C A B − = −

2sin cos cos2 2 2C A B A B + − = −

cos cos cos2 2 2C A B A B + + = − =

∵

2sin 2sin sin 4sin sin sin2 2 2 2 2 2C A B A B C = − = −

∴ Statement (c) is incorrect. (d) When ,A B C π+ + = we have ( ) ( )sin 2 sin 2 sin 2 2sin cos 2sin cosA B C A B A B C C+ + = + − + ( )2sin cos 2sin cosC A B C C= − + ( ) ( )sin sin sinA B C Cπ+ = − = ∵ ( ) ( ){ }2sin cos cosC A B A B= − − + ( ){ } ( )cos cos cosC A B A Bπ = − + = − + ∵ { }2sin 2sin sinC A B= 4sin sin sinA B C= ∴ Statement (d) is correct. 159. Ans. (d) sin sin sin3 3π π

θ θ θ − +

2 2sin sin sin3πθ θ

= −

2 2 33 1 1sin sin sin (3 4sin ) (3sin 4sin )4 4 4θ θ θ θ θ θ = − = − = −

1 sin 34 θ= 160. Ans. (b), Given, 1sin 10φ = 2 1 3cos 1 sin 1 .10 10φ φ⇒ = − = − =



1sin 110tan 3cos 310φ

φφ

∴ = = =

[ ]Note:sin , cos , tan areall +vesince lies in1st quad.φ φ φ φ

and so, 22212.2 tan 333tan 2 81 tan 411 93

φφ

φ

= = = =

− −

, Also, 1tan .7θ =

( )

251 3tan tan 2 287 4tan 2 11 3 251 tan tan 2 1 .7 4 28θ φ

θ φθ φ

+ + ∴ + = = = =

− −

( )tan 2 tan 45θ φ⇒ + = ° 2 45 .θ φ⇒ + = ° 161. Ans.(c), ( )cosec sin cos cos sinA B C B C+ ( )cosec sinA B C= ⋅ + ( )( )cosec sinA Aπ= − [ ]A B C B C Aπ π+ + = ⇒ + = −∵ cosec .sin 1A A= = 162. Ans. (c), ( ) ( )cos6 cos 4 cos 2 1θ θ θ+ + + 22cos5 cos 2cosθ θ θ= + { } { }2cos cos cos5 2cos 2cos3 cos 2θ θ θ θ θ θ= + = 4cos cos 2 cos3 .θ θ θ= 163. Ans. (c), ( ) ( )2 2 2 2sin cosec cos sec tan cotkα α α α α α+ + + = + + 2 2 2 2 2 2sin cosec 2 cos sec 2 tan cotkα α α α α α⇒ + + + + + = + + ( )2 2 2 2 2 24 sin cos cos sec tan cotec kα α α α α α⇒ + + + + = + + ( ) ( )2 2 2 25 cot 1 tan 1 tan cotkα α α α⇒ + + + + = + + 2 2 2 27 tan cot tan cotkα α α α⇒ + + = + + 7.k⇒ = 164. Ans. (d), sin 70 cos 40cos 70 sin 40°+ °

° + ° ( )

( )sin 70 sin 90 50cos 70 sin 90 50° + °− °

=° + °− °

sin 70 sin 50cos 70 cos50° + °

=° + °

( ) ( )sin cos 90 and cos sin 90θ θ θ θ= °− = °− ∵ 2sin 60 cos102cos60 cos10° °

=° °

sin sin 2sin cos2 2and cos cos 2cos cos2 2

C D C DC D

C D C DC D

+ − + = + − + =

∵ tan 60 3.= ° = 165. Ans.(a), 2 2cos sin4 4π π

θ θ + − −

or use sin cos cos4 2 4 4π π π πθ θ θ

− = − − = +

cos .cos4 4 4 4π π π π

θ θ θ θ = + + − + − −

( ) ( )2 2cos sin cos .cosA B A B A B − = + − ∵ cos .cos 2 02π θ= = cos 02π =

∵ 166. Ans. (b), Let ( )2 27 7sin cos 1 cos cos4 4P θ θ θ θ= − + = + − − ( )211 cos cos4 θ θ= − +


29


211 1 1cos cos4 4 4θ θ = + − + +

( )2211 1 1 1cos 2 cos4 4 2 2θ θ

= + − + +

213 cos .2θ = − +

Maximum value of P is obtained when 21cos 02θ + =

i.e.

When 1cos 2θ = − and max. value of 3 0 3.P = − = Minimum value of P is obtained when 21cos 2θ +

has the maximum value i.e. when cos 1θ = and

So the minimum value of 2 21 3 33 1 32 2 4P = − + = − =

∴ Required ratio Max.valueof 3 4.Min.valueof 3 / 4P

P= = =

167. Ans. (b), We have, 5sin 3sin 33πθ θ + + +

5sin 3 sin cos cos sin 33 3π πθ θ θ = + + +

3 3 35sin sin cos 32 2θ θ θ= + + + 13 3 3sin cos 32 2θ θ= + + Now, the maximum value of 13 3 3sin cos2 2θ θ+ is 2213 3 3 169 27 196 49 72 2 4 4 4 + = + = = =

2 2We know that the max.valueof sin cos isa b a bθ θ + +

∴ The maximum value of 13 3 3sin cos 32 2θ θ+ + is 10 Thus, the maximum value of 5sin 3sin 33πθ θ + + +

is 10.

168. Ans. (c), tan 20 . tan 40 . tan 60 . tan80° ° ° ° 3.tan 20 . tan 40 .tan 80 .= ° ° ° ( ) ( )3.tan 20 .tan 60 20 .tan 60 20= ° °− ° ° + ° ( )3.tan 3 20= × ° ( ) ( )Using the formula tan . tan 60 tan 60 tan 3θ θ θ θ° − ⋅ ° + = 3. tan 60 3. 3 3.= ° = = 169. Ans. (d) cos cos cos3 3π π

θ θ θ − +

2 2cos cos sin3πθ θ = −

2 21 1cos sin cos {1 4sin }4 4θ θ θ θ = − = −

2 21 1cos {1 4(1 cos )} cos (4cos 3)4 4θ θ θ θ= − − = − 31 1(4cos 3cos ) cos34 4θ θ θ= − = . 170. Ans. (b), sin163 cos347 sin 73 sin167° ° + ° ° ( ) ( ) ( ) ( )sin 180 17 cos 360 13 sin 90 17 sin 180 13= °− ° °− ° + °− ° ° − ° sin17 cos13 cos17 sin13= ° °+ ° ° ( ) 1sin 17 13 sin 30 .2= °+ ° = ° =



171. Ans. (b) 2 sin 2cos4 4π πθ θ + + + −

12 2 (sin cos )2 θ θ = + + +

12 2 . 2 sin cos cos sin4 42 π πθ θ = + + +

2 (2 1).sin 2 3sin4 4π πθ θ = + + + = + +

.

Clearly, the maximum value of sin 4πθ +

is 1. ∴ The maximum value of 2 sin 2cos4 4π π

θ θ + + + −

2 3= + .

22tan 1sin sec 1m

m

θθ

θ−

∴ = =+

172. Ans. (d), 4A B

π− = 4A B

π⇒ = +

1 tantan 1 tan BAB

+∴ =

− 21 tan 1 tanA

B⇒ + =

− ( )( )1 tan 1 tan 2.A B⇒ + − =

173. Ans. (a), sin 600 cos330 cos120 sin150° °+ ° ° ( ) ( ) ( ) ( )sin 360 240 cos 360 30 cos 180 60 sin 180 30= °+ ° ° − ° + °− ° °− ° ( )sin 240 cos30 cos60 sin 30= ° ° + − ° ° ( )sin 180 60 cos30 cos 60 sin 30= °+ ° °− ° ° sin 60 cos30 cos 60 sin 30= − ° °− ° ° ( )sin 60 cos30 cos60 sin 30= − ° ° + ° ° ( )sin 60 30 sin 90 1.= − °+ ° = − ° = − 174. Ans. (c), sin 2 sin 2 sin 2A B C+ + ( ) ( )2sin cos sin 2A B A B C= + − + ( )2sin cos 2sin cos2 C A B C C

π = − − +

2A B Cπ + + =

∵ ( )2cos cos 2sin cosC A B C C= − + ( ) ( )2cos cos sin 2C A B A B

π = − + − +

( ) ( )2cos cos cosC A B A B= + + − [ ]2cos 2cos cos 4cos cos cos .C A B A B C= = 175. Ans. (a) sin12 .sin 48 .sin 54° ° ° 1 .(2sin12 .sin 48 ).sin 542= ° ° °

1 .(cos36 cos 60 ).sin 54.2= °− ° 1 5 1 1 5 1. .2 4 2 4 + +

= −

1 5 1 5 1 4 1.2 4 4 32 8 − += = =

.

176. Ans. (b), Let 2 2cos cos cos3 3x y z

kπ πθ θ θ

= = = + −

Then, 2 2cos , cos , cos3 3x k y k z k

π πθ θ θ = = + = −

2 2cos cos cos3 3x y z kπ π

θ θ θ ∴ + + = + + + −

2cos 2cos cos 3k

πθ θ = +


31


{ }cos cosk θ θ= − 2 1cos cos cos3 3 3 2π π ππ

= − = − = − ∵

. 0 0.k= = 177. Ans. (c),

( ) ( )

2 2 2 2sin sin sin sin1 1sin cos sin cos sin 2 sin 22 2A B A B

A A B BA B

− −=

− −

( )( ) ( ){ }

2 2sin sin1 2cos sin2A B

A B A B

−=

+ − sin sin 2cos sin2 2C D C D

C D + − − = ∵

( ) ( )( ) ( )

( )( )

( )sin sin sin tancos sin cosA B A B A B

A BA B A B A B

+ − += = = +

+ − +

178. Ans. (c), 85 35 85 352cos sinsin 85 sin 35 2 2cos65 cos65°+ ° ° − °

°− ° =° °

2cos60 sin 25sin 25° ⋅ °

=°

( )cos65 sin 90 65 sin 25° = °− ° = ° ∵ 12cos60 2 12= ° = × = 179. Ans. (b), ( ) ( ) ( )

( ) ( ) ( )sin 660 tan 1050 sec 420cos 225 cosec 315 cos 510− ° ° − °

° ° ° ( ) ( ) ( )

( ) ( ) ( )sin 660 tan 1050 sec 420cos 225 cosec 315 cos 510° ° °

=° ° °

( ) ( ) ( )

( ) ( ) ( )sin 720 60 tan 1080 30 sec 360 60cos 180 45 cosec 360 45 cos 360 150− °− ° ° − ° ° + °

=°+ ° °− ° °+ °

{ } { }{ }{ }{ } { }

sin 60 tan 30 sec60cos 45 cosec45 cos150− − ° ⋅ − ° °=

− ° − ° ⋅ °

sin 60 tan 30 sec60cos 45 . cosec 45 .cos30° ° ⋅ °=

° ° ° [ ]cos150 cos30° = − °∵

3 1. .2 22 31 3 3. 2. 22= =

180. Ans. (d), 532sin sin2 2A A ( )16 cos 2 cos3A A= − ( ) ( )Using : 2sin sin cos cosA B A B A B= − − + ( ) ( )2 316. 2cos 1 4cos 3cosA A A = − − − 2 33 3 316. 2. 1 4. 3.4 4 4 = − − −

9 27 9 1 9 1116. 1 16. 16. 118 16 4 8 16 16 = − − − = + = =

181. Ans. (d), Let 2 24 tan and 9cotx yθ θ= = Then, A.M. G.M.≥ 2x y

xy+

⇒ ≥ [Where &x y both are non-negative] 2 2 2 24 tan 9cot 4 tan .9cot2θ θ

θ θ+

⇒ ≥ 2 24 tan 9cot 12θ θ⇒ + ≥ ∴ The minimum value of 2 24 tan 9cotθ θ+ is 12. 182. Ans. (d), Taking each choice separately (a) ( ){ } ( )sin sin sin sin cos cos sinA B C B C B C B Cπ= − + = + = + , ∴ Option (a) is true.



(b) ( ){ } ( )cos cos cos sin sin cos cos .C A B A B A B A Bπ= − + = − + = − ∴ Option (b) is true. For any angles , andA B Cwe have (c) ( ) tan tan tan tan tan tantan 1 tan tan tan tan tan tanA B C A B C

A B CA B B C C A

+ + −+ + =

− − − …(i)

In a triangle, we have A B C π+ + = , ∴ Equation (i) is reduced to tan tan tan tan tan tantan 1 tan tan tan tan tan tanA B C A B C

A B B C C Aπ

+ + −=

− − − 0 tan tan tan tan tan tanA B C A B C⇒ = + + −

tan tan tan tan . tan . tanA B C A B C⇒ + + = , ∴ Option (c) is true. (d) However. Option (d) is false for a triangle (It can be shown to be false e.g. for 60A B C= = = ° it is false) . 183. Ans. (c), cos cos cos 0x y α+ + = cos cos cosx y α⇒ + = − …(i) sin sin sin 0x y α+ + = sin sin sinx y α⇒ + = − …(ii) Dividing (i) by (ii) we get cos cos cossin sin sinx y

x y

αα

+ −=

+ − 2cos cos2 2 cot2sin cos2 2

x y x y

x y x yα

+ − ⇒ =

+ −

cot cot .2x y

α+ ⇒ =

184. Ans. (b), 2sec θ has a value greater than or equal to 1 i.e., 2sec 1θ ≥ ( )24 1xy

x y⇒ ≥

+

( )2 4x y xy⇒ + ≤ ( )2 4 0x y xy⇒ + − ≤ ( )2 0x y⇒ − ≤ ( )2 0x y⇒ − = ( )2 never be vex y − − ∵ .x y⇒ = Also, when x y= and 0,x = then the value of 2sec θ becomes undefined Hence,

( )2 24sec xy

x yθ =

+ is true if and only if , 0.x y x= ≠

185. Ans.(b), 180A B C+ + = ° 902 2 2A B C⇒ + = °− tan tan 902 2 2A B C ⇒ + = °−

tan tan2 2 cot 21 tan tan2 2A B

C

A B

+⇒ =

− tan tan 12 21 tan tan tan2 2 2

A B

A B C

+⇒ =

−

tan tan tan tan 1 tan tan2 2 2 2 2 2A C B C A B⇒ + = − tan tan tan tan tan tan 12 2 2 2 2 2A B B C C A

⇒ + + = tan tan 1.2 2A B

= =∑ 186. Ans. (a), tan . tan .tan tan 33 3 k

π πθ θ θ θ + − + =

tan tantan tan 33tan . . tan 31 tan tan 1 tan tan3 3k

ππ θθθ θ

π πθ θ

− ++ ⇒ = − − −


33


3 tan 3 tantan . tan 31 3 tan 1 3 tan kθ θ

θ θθ θ

+ − +⇒ = − +

22tan 3tan tan 31 3tan k

θθ θ

θ −

⇒ = −

323tan tan tan 31 3tan k

θ θθ

θ −

⇒ − = − tan 3 tan 3kθ θ⇒ − = 1.k⇒ = −

Alternatively : We know that tan . tan .tan tan 33 3π πθ θ θ θ − + =

tan . tan . tan tan 33 3π πθ θ θ θ ∴ + − + = −

and so, 1k = −

187. Ans. (b), sin sinA n B= sinsin 1A n

B⇒ = sin sin 1sin sin 1A B n

A B n

− −⇒ =

+ + [By Componendo-dividendo]

2cos sin 12 2 12sin cos2 2A B A B

n

A B A B n

+ − − ⇒ =

+ − +

tan 12 1tan 2A B

n

A B n

− − ⇒ =

+ +

1 tan tan1 2 2n A B A B

n

− + − ⇒ = + .

188. Ans. (d), cos 2 cos 2 cos 2A B C+ + ( ) ( ){ } ( )22cos cos 2cos 1A B A B C= + − + − ( ) ( ) 22cos 180 cos 2cos 1C A B C= °− − + − [ ]180A B C+ + = °∵ ( ) 22cos cos 2cos 1C A B C= − − + − ( ){ }2cos cos cos 1C A B C= − ⋅ − − − ( ) ( )( ){ }1 2cos . cos cos 180C A B A B= − − − − °− + [ ]180A B C+ + = °∵ ( ) ( ){ }1 2cos . cos cosC A B A B= − − − + + { }1 2cos . 2cos cos 1 4cos cos cosC A B A B C= − − = − − 189. Ans. (c), Since 0, ,2πα ∈

so tan 0α ≥

Now, if 22 2tanand ,A x x Bx x

α= + =

+ then we have .2A B

A B+

≥ [ ]. . .AM GM≥∵ 1/ 22 22 22 2

1 tan tan.2 x x x xx x x x

α α ⇒ + + ≥ +

+ + 22 2tan 2 tan .x x

x x

αα⇒ + + ≥

+

190. Ans. (b), 5 1sin18 4−° = 2 5 1 5 1sin 18 .4 4 − −⇒ ° =

6 2 5 3 5 .16 8− −= =

5 1cos36 4+° = 2 5 1 5 1cos 36 .4 4 + +⇒ ° =

6 2 5 3 5 .16 8+ += =

∴ If 2 2sin 18 and cos 36° °are the roots of a quadratic equation, then Sum of the roots 3 5 3 5 6 38 8 8 4− +

= + = = and product of the roots 3 5 3 5 4 1. .8 8 64 16 − += = =

∴ Required quadratic equation is 2 3 1 04 16x x − + =

216 12 1 0.x x⇒ − + = 191. Ans. (b), 180A B C+ + = ° 902 2 2A B C

⇒ + + = ° tan tan 902 2 2A B C ⇒ + + = ° = ∞

1 321S S

S

−⇒ = ∞

−

Where, 1 tan tan tan tan2 2 2 2A B C AS = + + = ∑



2 tan tan2 2A BS

= ⋅

∑ and 3 tan tan tan2 2 2A B CS = and 1 3

21S S

S

−= ∞

− 21 0S⇒ − =

2 1S⇒ = tan tan 12 2A B⇒ =∑ [Note : Remember this result directly]

192. Ans. (a) sin sin sin sin 0α β α β= ⇒ − = 2cos sin 02 2α β α β+ − ⇒ =

…(i)

And cos cos cos cos 0α β α β= ⇒ − = 2sin sin 02 2α β α β+ − ⇒ − =

…(ii)

From (i) and (ii) we get: sin 02α β− =

cos 0, and sin 0 both simultanevely is not possible2 2α β α β+ + = =

∵

i.e., the two conditions are satisfied only if sin 02α β− =

. 193. Ans. (b) We have: cos 2 cos 2 cos 2A B C+ + 22cos( )cos( ) (1 2sin )A B A B C= + − + − 22sin cos( ) 1 2sinC A B C= − − + − 1 2sin {cos( ) sin }C A B C= − − + 1 2sin {cos( ) cos( )}C A B A B= − − − + 1 2sin {2sin sin }C A B= − 1 4sin sin sinA B C= − cos 2 cos 2 cos 2 4sin sin sin 1A B C A B C⇒ + + + = . 194. Ans. (b), ( )

( )1 cos8sec8 1 cos 4.sec 4 1 cos8 1 cos 4AA A

A A A

−−=

− − 2

22sin 4 cos 4.cos8 2sin 2A A

A A= 21 cos 2 2sinθ θ − = ∵

( )22sin 4 . cos 4 .sin 4cos8 . 2sin 2A A A

A A= ( )

( )sin8 . 2sin 2 cos 2 tan 8 .cos8 . 2sin 2 .sin 2 tan 2A A A A

A A A A= =

195. Ans. (b), sin 36 sin 72 sin108 sin144° ° ° ° sin 36 cos18 cos18 sin 36= ° ° ° ° ( ) ( )

( )sin 72 sin 90 18 cos18 , sin108 sin 90 18 cos18 and sin144sin 180 36 sin 36° = °− ° = ° ° = °+ ° = ° °

= ° − ° = °

∵ ( ) ( )1 11 cos 72 . 1 cos362 2= − ° + ° ( )

( )

22

1sin 1 cos 22 1and cos 1 cos 22A A

A A

= − = +

∵ 1 5 1 5 11 14 4 4 − +

= − +

( )( )1 1 55 5 5 5 .20 .64 64 16= − + = = 196. Ans.(d), 2 2 2cos cos3 3x x

π π + + +

1 1 21 cos 2 1 cos 22 3 2 3x xπ π = + + + + +

1 2 41 cos 2 cos 22 3 3x xπ π = + + + +

( )11 2cos 2 cos2 3x

ππ

= + +

( )11 cos 2 .2 x= − Now, ( )11 cos 22 x− attains a maximum value, When ( )1 cos 22 x has the minimum value i.e. when ( )cos 2x has the minimum value i.e., when ( )cos 2 1.x = −


35


∴ Maximum value of ( ) ( )1 1 31 cos 2 1 . 12 2 2x− = − − = and so, the maximum value of 2 2 2 3cos cos is3 3 2x x

π π + + +

. 197. Ans. (a), Given, secx h a θ= + sec x h

aθ

−⇒ = cos a

x hθ⇒ =

−

( )

22 2cos a

x hθ⇒ =

− and cosecy k b θ= + cosec y k

bθ

−⇒ = sin b

y kθ⇒ =

−

( )

22 2sin b

y kθ⇒ =

−

Now, 2 2cos sin 1θ θ+ = ( ) ( )

2 22 2 1a b

x h y k⇒ =

− −

198. Ans. (a), ( )2 2 24cos cos cos3 3x x xπ π + −

( )214. cos 34 x= 1Using formula cos .cos cos cos33 3 4π πθ θ θ θ

− + =

( )2cos 3x= and Clearly, ( )21 cos 3 1x− ≤ ≤ [ ]We know that 1 cos 1θ− ≤ ≤ ⇒ The extreme values of the given expression are 1− and 1. 199. Ans. (a), ( ) ( )4 6 4 636 sin 5 sin 4 sin sin 32 2π π

π θ θ θ π θ + + + − + + +

{ } { }4 4 6 66 sin cos 4 cos sinθ θ θ θ= + − + ( )

( )

sin 5 sin , sin cos ,23sin cos , sin 3 sin2π

π θ θ θ θ

πθ θ π θ θ

+ = − + = − + = − + = −

∵ ( ){ }22 2 2 26 sin cos 2sin cosθ θ θ θ= + − ( ) ( ){ }32 2 2 2 2 24 cos sin 3sin cos cos sinθ θ θ θ θ θ− + − + { } { }2 2 2 26 1 2sin cos 4 1 3sin cos 2.θ θ θ θ= − − − = 200. Ans. (c), tan cos tanβ θ α= tan costan 1β θ

α⇒ =

tan tan 1 costan tan 1 cosα β θα β θ

− −⇒ =

+ + [By Componendo and Dividendo]

22

sin sin 2sincos cos 2sin sin 2coscos cos 2α β θα βα β θα β

−⇒ =

+ 2sin cos cos sin tansin cos cos sin 2α β α β θ

α β α β−

⇒ =+

( )( )

2sin tan .sin 2α β θα β

−⇒ =

+

201. Ans. (a), 2 2 2 23 4cos cos cos 108 cos 1445 5π π+ = °+ ° ( ) ( )2 2cos 90 18 cos 180 36= °+ ° + °− °

( ) ( )2 2 2 2sin18 cos36 sin 18 cos 36= − ° + − ° = ° + ° 2 25 1 5 1 6 2 5 6 2 54 4 16 16 − + − += + = +

12 3 .16 4= = 202. Ans. (d), ( )

( )cos 1sin 1p

p

α βα β

− +=

+ − ( ) ( ) ( ) ( )1 cos 1 sinp pα β α β⇒ − − = + +



( ) ( ) ( ) ( )cos sin sin cosp α β α β α β α β⇒ − − + = + + − ( ) ( )( ) ( )

sin coscos sinpα β α βα β α β

+ + −⇒ =

− − +

sin cos cos sin cos cos sin sincos cos sin sin sin cos cos sinpα β α β α β α βα β α β α β α β

+ + +⇒ =

+ − −

tan tan 1 tan tan1 tan tan tan tanpα β α β

α β α β+ + +

⇒ =+ − −

[ ]Dividing Nr. and Dr. by cos cosα β ( ) ( )

( ) ( )1 tan 1 tan 1 tan 1 tan1 tan 1 tan 1 tan 1 tanp

α β α βα β α β

+ + + + ⇒ = = − − − − tan .tan .4 4p

π πα β ⇒ = + +

203. Ans. (c), sin 4 cos 2 cos 4 sin 2A A A A− = − sin 4 sin 2 cos 4 cos 2A A A A⇒ + = + 2sin 3 cos 2cos3 cosA A A A⇒ = sin 3 cos3A A⇒ = 0 cos 04A A

π < < ⇒ ≠ ∵

tan 3 1 tan 45A⇒ = = ° 3 45A⇒ = ° 30 0 3 and so tan 3 1 tan 3 tan 454 4A A A A

π π < < ⇒ < < = ⇒ = ° ∵

15A⇒ = ° 4 60A⇒ = ° tan 4 tan 60 3.A⇒ = ° = 204. Ans. (b), 2 3cos cos2 2B C A A B+ + − +

3cos cos2 2A C A A Bπ − + + − = +

[ ]B C Aπ+ = −∵

2cos cos2 2C A A Bπ + + − = +

( )cos cos2 2B A A Bπ π+ − + − = +

[ ]C A Bπ+ = −∵ 2cos cos2 2A B A Bπ + − − ⇒ +

cos cos2 2A B A B

π− − = + +

cos cos 0.2 2A B A B− − = − + =

205. Ans. (b), We have 2 2cot 36 cot 72° ° 2 2

2 2cos 36 cos 72sin 36 sin 72° °=

° °

( )( )( )( )

2 22 2

2cos 36 2cos 722sin 36 2sin 72° °=

° ° ( )( )

( )( )1 cos72 1 cos1441 cos72 1 cos144+ ° + °

=− ° − °

( )( )

( ) ( )1 cos 72 1 cos361 cos 72 1 cos36+ ° − °

=− ° + °

( )cos144 cos 180 36 cos36° = ° − ° = − ° ∵ ( )( )

( )( )

5 1 5 11̀ 1 3 5 3 54 45 1 5 1 5 5 5 51 14 4` − +

+ − + − = = − + − +

− +

9 5 4 125 5 20 5−= = =

−, Hence, option (b) is correct.

Alternatively: Given 2 22 2cos 36 cos 72sin 36 sin 72° °

=° °

2 2

25 1 5 1 1 4 14 4 4 5100 20 8010 2 5 10 2 5 164 4 + − × = = = = − − +

206. Ans. (c) We have: tan tanA B x− = …(i) cot cotB A y− = …(ii)


37


AB

C 3 1−

3 1+

θ 712 2

Dividing (i) by (ii) we get: tan tancot cotA B x

B A y

−=

−

tan tan tan tan1 1tan tanA B x x

A By y

B A

−⇒ = ⇒ =

− …(iii)

Now, tan tantan( ) 1 tan tan 1A B xA B

xA B

y

−− = =

+ + [Using(i) & (iii)]

tan( ) xyA B

x y⇒ − =

+ ( ) 1 1cot x y

A Bxy y x

+⇒ − = = +

207. Ans. (b), ∵ 3 1 Perpendiculartan Base3 1 BC

AAB

−= = =

+

∴ ( ) ( )2 2

3 1cos 3 1 3 1AB

AAC

+= =

− + + 3 12 2+=

208. Ans. (b), ∵ 1tan 7θ = PerpendicularBase= ∴ 1 1sin ,7 1 2 2θ = =

+ 7cos 2 2θ =

∴ cosec 2 2θ = and 2 2sec 7θ = ∴ ( )

( )

2 22 2 8 8 / 7cosec seccosec sec 8 8 / 7θ θθ θ

−−=

+ + 48 364 4= =

209. Ans. (d), Given, tan cot 2θ θ+ = ⇒ 1tan 2tanθθ

+ = ⇒ 2tan 1 2 tanθ θ+ = ⇒ 2tan 2 tan 1 0θ θ− + = ⇒ ( )2tan 1 0θ − = ⇒ tan 1θ = ∴ 45θ = ° ∴ 1 1 2sin cos sin 45 cos 45 22 2 2θ θ+ = °+ ° = + = = 210. Ans. (a), ∵ 2cos tan 2cos tan 1 0A B A B⋅ − − + = ∴ ( ) ( )2cos tan 1 tan 1 0A B B− − − = ∴ ( )( )2cos 1 tan 1 0A B− − = ∴ 1cos cos60 ,2A = = ° tan 1 tan 45B = = ° ∴ 60 ,A = ° 45B = ° 211. Ans. (c), Given, 3 3 3 3sin cos sin cossin cos sin cosθ θ θ θ

θ θ θ θ+ −

++ −

( )( ) ( ) ( )

( )

2 2 2 2sin cos sin sin cos cos sin cos sin sin cos cossin cos sin cosθ θ θ θ θ θ θ θ θ θ θ θ

θ θ θ θ

+ − ⋅ + − + ⋅ += +

+ −

1 sin cos 1 sin cosθ θ θ θ= − ⋅ + + ⋅ 2= 212. Ans. (d), ( ) ( )sec tan 1 sec tan 1θ θ θ θ+ − − + ( ) ( )sec tan 1 sec tan 1θ θ θ θ= + − − − ( )22sec tan 1θ θ = − − ( )2 2sec tan 2 tan 1θ θ θ= − − + 2 2sec tan 2 tan 1θ θ θ= − + − 1 2 tan 1 2 tanθ θ= + − =



θ 2 2q p−

P

q

213. Ans. (d), ( ) ( )2 2sin cosec cos secθ θ θ θ+ + + ( ) ( ) ( ) ( )2 2 2 2sin cosec 2 sin cosec cos sec 2 cos secθ θ θ θ θ θ θ θ= + + + + + ( ) ( ) ( )2 2 2 2sin cos 2 1 2 1 sec cosecθ θ θ θ= + + + + +

( ) ( ) ( )2 21 2 2 1 tan 1 cotθ θ= + + + + + + 2 2tan cot 7θ θ= + + , ∴ 7k = 214. Ans. (a), ( )4 4 2 2sin cos sin cosθ θ θ θ+ + ( )22 2 2 2 2 2sin cos 2sin cos sin cosθ θ θ θ θ θ = + − +

( )2 2 2 2 21 2sin cos sin cosθ θ θ θ= − + 2 2 21 sin cos 1 uθ θ= − = − , ∴ sin cosu θ θ= ⋅ 215. Ans. (d), ( )( ) ( )( )2 2sin cos 1 sin cos sin cos sin cos sin cosθ θ θ θ θ θ θ θ θ θ+ − = + + − ( )( )2 2sin cos sin sin cos cosθ θ θ θ θ θ= + − + 3 3sin cosθ θ= + sin cosn nθ θ= + ∴ 3n = 216. Ans. (a), ∵ 2 2cos sinp q pθ θ⋅ = − ⋅ ∴ 2 2

Perpendicular tan Basep

q pθ = =

−

∴ Perpendicularsin Hypoteneous θ = ( )2 2 2p p

qp q p= =

+ −

∴ sinq pθ⋅ = 217. Ans. (a), ∵ sin 3 cosθ θ= ⋅ ∴ tan 3θ = ∴ 60θ = ° ∴ ( )2 sin cos 1θ θ+ − ( )sin 60 cos60 1= °+ ° − 3 12 1 3 1 1 32 2

= + − = + − =

218. Ans. (d), sec cos 2θ θ+ = 1 cos 2cos θ

θ⇒ + = 2cos 2cos 1 0θ θ⇒ − + = ⇒ ( )2cos 1 0θ − =

cos 1θ⇒ = , ∴ 2 4sec sec 1 1θ θ− = − = 0, Hence, option (d) is correct. 219. Ans. (a), tan cot 2θ θ+ = 1tan 2tanθ

θ⇒ + = ( )2tan 1 0θ⇒ − = tan 1θ⇒ =

∴ 2 3tan tan 1 1 0θ θ− = − = , Also, in option (a) 3 2cot cot 1 1 0θ θ− = − = 220. Ans. (c), ∵ 1 sin sin cos cos 0α β α β+ ⋅ − ⋅ = ∴ cos cos sin sin 1α β α β⋅ − ⋅ = ∴ ( )cos 1α β+ = ∴ 2nα β π+ = ∴ 2nβ π α= − ∴

( )tan tan tantan cot 1tan tan 2 tann

α α αα β

β π α α⋅ = = = = −

− −

221. Ans. (b), ∵ A B C π+ + = ∴ 2 2 2B C Aπ+

= − and 2 2 2A B Cπ+= − ∴ tan tan cot2 2 2 2B C A Aπ+ = − =

cos cos sin2 2 2 2A B C Cπ+ = − =

∴ given expression tan tan cos cosec2 2 2 2A B C A B C+ += ⋅ + ⋅

tan cot sin cosec2 2 2 2A C CΑ = ⋅ + ⋅

1 1 2= + = 222. Ans. (c), 2 2 2x y z+ + 2 2 2 2 2 2 2 2cos cos cos sin sinr A B r A B r A= ⋅ ⋅ + ⋅ ⋅ + ⋅ ( )2 2 2 2 2 2cos cos sin sinr A B B r A= ⋅ + + ⋅ 2 2 2 2cos sinr A r A= + ( )2 2 2cos sinr A A= + 2r=


39


223. Ans. (a), ∵ 2 2cos sinx a bθ θ= ⋅ + ⋅ ∴ ( )2 2cos sinx a a b aθ θ− = ⋅ + ⋅ − ( )2 2sin 1 cosb aθ θ= ⋅ − − 2 2sin sinb aθ θ= ⋅ − ⋅ ( ) 2sinb a θ= − ⋅ and ( )2 2cos sinb x b a bθ θ− = − ⋅ + ⋅ ( )2 21 sin cosb aθ θ= − − ⋅ 2 2cos cosb aθ θ= ⋅ − ⋅ ( ) 2cosb a θ= − ⋅ ∴ ( )( ) ( ) ( )2 2sin cosx a b x b a b aθ θ− − = − ⋅ ⋅ − ⋅ ( )2 2 2sin cosb a θ θ= − ⋅ ⋅ ( )2 2 2sin cosa b θ θ= − ⋅ ⋅ ( )( ) 2 2 2sin cosx a b x c θ θ− − = ⋅∵ ∴ ( )22c a b= − ∴ c a b= − or b a− 224. Ans. (b), Given, ( ) ( )cos cosx A B A B= + ⋅ − and ( ) ( )sin siny A B A B= + ⋅ − ∴ ( ) ( ) ( ) ( )cos cos sin sinx y A B A B A B A B− = + ⋅ − − + ⋅ − ( ) ( )cos A B A B= + + − cos 2A= [∵ cos cos sin sinα β α β⋅ − ⋅ ( )cos α β= + ] 225. Ans. (a), Given, sin cos3 3x A B

π π = + ⋅ +

and cos sin3 3y A Bπ π = + ⋅ +

∴ sin cos cos sin3 3 3 3x y A B A Bπ π π π − = + ⋅ + − + ⋅ +

( )sin sin3 3A B A Bπ π = + − + = −

[∵ sin cos cos sinα β α β⋅ − ⋅ ( )sin α β= + ]

226. Ans. (d), 1 tan tan 2AA + ⋅

sinsin 21 cos cos 2A

A

AA= + ⋅ cos cos sin sin2 2cos cos 2

A AA A

AA

⋅ + ⋅=

⋅

cos cos2 2cos cos cos cos2 2A AA

A AA A

− = =

⋅ ⋅ 1 seccos A

A= =

227. Ans. (b), ∵ ( )cos cos sinx A B A B⋅ ⋅ = − ∴ ( )sincos cosA Bx

A B

−=

⋅ sin cos cos sincos cosA B A B

A B

⋅ − ⋅=

⋅

∴ tan tanx A B= − , similarly, tan tany B C= − and tan tanz C A= − ∴ tan tan tan tan tan tan 0x y z A B B C C A+ + = − + − + − = 228. Ans. (c), Given, ( )sin cosx B C A= − ⋅ , ( )sin cos ,y C A B= − ⋅ and ( )sin cosz A B C= − ⋅ ( )sin cosx B C A⇒ = − ⋅ ( )sin cos cos sin cosB C B C A= ⋅ − ⋅ ⋅ cos cos cos cos cos cosA B C A B C= ⋅ ⋅ − ⋅ ⋅ , Similarly, cos cos sin cos cos siny A B C B C A= ⋅ ⋅ − ⋅ ⋅ and sin cos cos cos cos sinz A B C B C A= ⋅ ⋅ − ⋅ ⋅ So, x y z+ + = cos cos cos cos cos cosA B C A B C⋅ ⋅ − ⋅ ⋅ + cos cos sin cos cos sinA B C B C A⋅ ⋅ − ⋅ ⋅ sin cos cos cos cos sinA B C B C A+ ⋅ ⋅ − ⋅ ⋅ 0= 229. Ans. (d), Given, ( ) ( )2 sin 60 cos 30x x⋅ + ° = − ° [ ]2 sin cos 60 cos sin 60 cos cos30 sin sin 30x x x x⇒ ⋅ °+ ⋅ ° = ⋅ ° + ⋅ ° ⇒ 1 32 sin cos2 2x x

⋅ + ⋅

3 1cos sin2 2x x= ⋅ + ⋅ ∴ sin 3 cosx x+ ⋅ ( )1 3 cos sin2 x x= ⋅ +



2sin 2 3 cos 3 cos sinx x x x⇒ + = + ⇒ sin 3 cosx x= − ⋅ ∴ tan 3x = − 230. Ans. (b), ∴ ( ) ( )cos cos 120 cos 120x x x+ °− + °+ 2 cos120 cos cosx x= ⋅ ° ⋅ + ( )2 cos 90 30 cos cosx x= ⋅ ° + ° ⋅ + ( )2 sin 30 cos cosx x= − ° ⋅ + 12 cos cos cos cos 02 x x x x

= − ⋅ + = − + =

231. Ans. (c), ( ) ( )tan 45 tan 45A A° + ⋅ ° − 1 tan 1 tan1 tan 1 tanA A

A A

+ −= ⋅

− + 1=

232. Ans. (d), tan 75 tan15° + ° ( ) ( )tan 45 30 tan 45 30= °+ ° + °− ° 1 tan 30 1 tan 301 tan 30 1 tan 30+ ° − °

= +− ° + °

1 11 1 3 1 3 13 31 1 3 1 3 11 13 3+ −

+ −= + = +

− +− + ( ) ( )2 23 1 3 1 8 43 1 2+ + −= = =

−

233. Ans. (b), ∵ tana x= ∴ 21 secc a x= + = , tanb y= ∴ 21 secd b y= + = ∴ ( )sincd x y⋅ + ( )( ) ( )sec sec sin cos cos sinx y x y x y= ⋅ + ⋅ sin cos cos sincos cosx y x y

x y

⋅ + ⋅=

⋅

tan tanx y= + a b= + 234. Ans. (c), ∵ 2 2cos sin4 2 4π π π

θ θ − = − −

2sin 2 4π π

θ = − +

2sin 4π θ = +

…(i) ∴ 2 2 2 2cos cos sin cos 14 4 4 4π π π π

θ θ θ θ − + + = + + + =

[Using (i)] 235. Ans. (c), ∵ 11cos cos cos12 12 12π π π

π = − = −

…(i) and 9 3 3cos cos cos2 12 12π π π

π = − = −

…(ii) ∴ 3 9 11cos cos cos cos12 12 12 12π π π π

+ + + = 3 3cos cos cos cos 0 cos12 12 12 12 2π π π π π+ − − = =

6cos 12π= Hence, option (c) is correct. 236. Ans. (a), ∵ 9cot cot tan20 2 20 20π π π π = − =

7 3 3cot cot tan20 2 20 20π π π π = − =

and 5cot cot20 4π π=

∴ Given 3 3cot tan cot tan tan 1 1 1 120 20 20 20 4π π π π π = ⋅ ⋅ ⋅ ⋅ = × × =

237. Ans. (b), ∵ 7sin sin sin8 8 8π π π

π = − =

and 5 3 3sin sin sin8 8 8π π ππ = − =

∴ Given Expression 2 2 32 sin sin8 8π π = +

But, sin sin cos8 2 8 8π π π π3 = − =

…(i)

∴ Given Expression 2 22 sin cos8 8π π = +

( )2 1 2= = [Using (i)]



238. Ans. (d), 1 sin cos1 sin cosθ θθ θ

+ ++ −

( )( )1 cos sin1 cos sinθ θ

θ θ+ +

=− +

22

2cos 2sin cos2 2 22sin 2sin cos2 2 2θ θ θ

θ θ θ

+ ⋅=

+ ⋅ cot 2θ=

239. Ans. (a), ∵ We know that, 221 tan cos 21 tan θ

θθ

−=

+

∴ ( )( )

( ) ( )221 tan 45 cos 2 45 cos 90 2 sin 21 tan 45 A

A A AA

− °−= °− = °− = + °−

240. Ans. (d), sin 3 cos3 sin 3 cos cos3 sinsin cos sin cosθ θ θ θ θ θ

θ θ θ θ⋅ − ⋅

− =⋅

( )sin 3 sin 2sin cos sin cosθ θ θ

θ θ θ θ−

= =⋅ ⋅

2sin cos 2sin cosθ θθ θ

⋅= =

⋅

241. Ans. (a), ( )sin 2sin 2 cos 2 sin 2 cos cos 2 sin sin secsin cos sin cos sin cos sin cosθ θθ θ θ θ θ θ θθ

θ θ θ θ θ θ θ θ−−

− = = = =⋅ ⋅ ⋅

242. Ans. (d), ( )3 3 2 24sin cos 4cos sin 4sin cos cos sinθ θ θ θ θ θ θ θ− = − ( ) ( )2 2sin cos cos 2θ θ θ= 2sin 2 cos 2θ θ= ⋅ sin 4θ= 243. Ans. (b), ( )2 4 2 21 8cos 8cos 1 8cos 1 cosθ θ θ θ− + = − − ( ) ( )2 2 2 21 8cos sin 1 2 4sin cosθ θ θ θ= − = − ( )21 2sin 2 cos 2 2θ θ= − = × cos 4θ= 244. Ans. (b), cos3 cos5 cos 7 cos15A A A A+ + + ( ) ( )cos15 cos3 cos7 cos5A A A A= + + + 2cos9 cos 6 2cos6 cosA A A A= ⋅ + ⋅ ( )2cos 6 cos9 cosA A A= ⋅ + ( )2cos 6 2cos5 cos 4A A A= ⋅ ⋅ 4 cos 4 cos5 cos 6A A A= ⋅ ⋅ ⋅ ⇒ 4 5 6 15m n p+ + = + + = 245. Ans. (c), Given, sin 40 cos 70 cos80k° − ° = ⋅ ° ( )sin 40 cos 90 20 cos80k⇒ °− °− ° = ⋅ ° sin 40 sin 20 cos80k⇒ °− ° = ⋅ ° 2cos30 sin10 cos80k⇒ °⋅ ° = ° 32 2 k

⇒ =

3k⇒ =

246. Ans. (c), ( )cos 20 cos100 cos140°+ ° + ° cos 20 2cos120 cos 20= °+ ° ° ( )cos 20 2cos 90 30 .cos 20= °+ °+ ° ° ( )cos 20 2 sin 30 cos 20= °+ − ° ⋅ ° 1cos 20 2 cos 202 = ° + − ⋅ °

cos 20 cos 20 0= °− ° =

247. Ans. (a), cos55 cos65 cos175° + ° + ° ( ) ( ) ( )cos 90 35 cos 90 25 cos 180 5= °− ° + °− ° + °− ° ( )sin 35 sin 25 cos5= °+ ° − ° = 2sin 30 cos5 cos5° ⋅ ° − ° 2 cos5 cos521 = ⋅ ° − °

cos5= ° cos5 0− ° =

248. Ans. (c), sin10 sin 20 sin 40 sin 50°+ ° + ° + ° ( ) ( )sin 50 sin10 sin 40 sin 20= °+ ° + °+ ° 2sin 30 cos 20 2sin 30 cos10= °⋅ ° + ° ° 12 cos 20 2 cos102 21 = ⋅ ° + ⋅ °

cos 20 cos10= °+ ° ( ) ( )cos 90 70 cos 90 80= °− ° + °− ° sin 70 sin80= °+ ° 249. Ans. (b), cos 21 sin 21sin 21 sin 21° − °

°+ °1 tan 21 tan 45 tan 211 tan 21 1 tan 45 tan 21− ° ° − °

= =+ ° + °⋅ °

[Divide N, D by cos 21° ] ( )tan 45 21 tan 24= °− ° = ° ( )tan 90 66 cot 66= °− ° = °



250. Ans. (a), sin 7 sin sin11 sin 3θ θ θ θ⋅ + ⋅ ( ) ( )1 12sin 7 sin 2sin11 sin 32 2θ θ θ θ= ⋅ + ( ) ( )1 1cos6 cos8 cos8 cos142 2θ θ θ θ= − + − ( )1 cos 6 cos142 θ θ= − ( )1 2sin10 sin 42 θ θ= ⋅ sin 4 sin10θ θ= ⋅ 251. Ans. (c), 2 3 4 516cos cos cos cos cos12 12 12 12 12π π π π π 3 5 2 44 4cos cos cos cos cos12 12 12 12 12π π π π π =

= ( )4 4 cos15 cos 45 cos 75 cos30 cos 60° ° ° ° ° = ( )4 4 cos15 cos(60 15 ) cos(60 15 ) cos30 cos 60° ° − ° ° + ° ° ° ( )4cos 3 15 cos30 cos60= × ° × °× ° [∵ We know that ( ) ( )4cos cos 60 cos 60 cos3θ θ θ θ⋅ ° − ⋅ ° + = ] 1 3 1 34 2 22 2= × × × = 32= 252. Ans. (c), Given expression, ( )sin 2 sin 2 sin 2A B C+ − ( ) ( )2sin cos sin 2A B A B C= + ⋅ − − ( )2sin cos 2sin cosC A B C C= ⋅ − − ⋅ ( )2sin cos cosC A B C= ⋅ − −

( ) ( )2sin cos cosC A B A B = ⋅ − − − + ( ) ( )2sin cos cosC A B A B= ⋅ + + − ( )2sin 2cos cosC A B= ⋅ ⋅ 4cos cos sinA B C= ⋅ ⋅ 253. Ans. (b), ( ) ( )sin 2 sin 2 sin 2A B A B+ + − ( ) ( ) ( ) ( )2sin cos 2sin cosA B A B A B A B= + ⋅ − + − ⋅ − ( ) ( ) ( )2cos sin sinA B A B A B= − ⋅ + + − ( ) ( )2cos 2sin cosA B A B= − ⋅ ⋅ ( )4sin cos cosA B A B= ⋅ ⋅ − 254. Ans. (c), 4 4tan cot 2A A+ + 4 4 2 2tan cot 2 tan cotA A A A= + + ( )22 2tan cotA A= + ( ) ( )2 2 2 2tan cot sec 1 cosec 1A A A A= + = − + − ( )2 2 2 21 1sec cosec 2 2cos sinA A

A A= + − = + −

2 22 2sin cos 2sin cosA A

A A

+= − 2 21 2sin cosA A

= − 2 2sec cosec 2A A= ⋅ − 255. Ans. (b), Given Expression 1 tan 1 tan1 tan 1 tanθ θ

θ θ+ −

= −− +

( ) ( )( )

2 221 tan 1 tan1 tanθ θ

θ

+ − −=

−

( ) 224 tan 2 tan2 1 tan1 tanθ θ

θθ = = −−

( ) ( )2 tan 2 tanm nθ θ= ⋅ = ⋅ ∴ 2, 2m n= = ∴ m n= 256. Ans. (a), Given expression, 2 2sin cos4 2 4 2π θ π θ + − +

cos 2 4 2π θ = − +

[∵ We know that 2 2cos sin cos 2x x x− = ] cos 2π θ = − +

( )sin sinθ θ= − − =

257. Ans. (c), Given, cot a

bθ = tan b

aθ⇒ = …(i)

∴ Given expression, 22 21 tan 2 tancos 2 sin 2 1 tan 1 tana b a bθ θ

θ θθ θ

− + = + + +



22 21 tan 2 tanWe know that cos 2 and sin 21 tan 1 tanθ θ

θ θθ θ

−= = + +

∵

22 2

1 21 1

b bb

a aa

b b

a a

− ⋅ = +

+ +

[Using (i)] ( )

( ) ( )

2 2 2 2 2 22 2 2 2

2 2a a b ab a a b b

a b a b

− + − + = =+ +

( )( )

2 22 2

a a ba

a b

+= =

+

258. Ans. (b) 3 52cos cos cos cos13 13 13 13π π π π9+ +

10 8 3 5cos cos cos cos13 13 13 13π π π π= + + +

3 5 3 5cos cos cos cos13 13 13 13π π π ππ π = − + − + +

3 5 3 5cos cos cos cos 013 13 13 13π π π π

= − − + + = 259. Ans. (c), Given that, radius 3r = m and arc 1d = m We know that, angle arc 1ardius 3= = rad 260. Ans. (a), ∵ We know that ( ) ( )cos 540 cos 3θ π θ°− = − cosθ= − and ( )sin 630 sin 7 cos2πθ θ θ ° − = − = −

∴ Given expression ( )cos cos cos cos 0θ θ θ θ= − − − = − + = 261. Ans. (a), Given 2 21 1 tan 25 1tan 40 cot 50 tan 50 2 tan 25 2 xx− ° −

= ° = ° = = =° °

Alternatively ∵ ( )tan155 tan 180 25 tan 25 x° = °− ° = − ° = − and ( )tan115 tan 90 25° = ° + °

( )1cot 25 tan 25−

= − ° =°

1x

= − ∴ Given expression

( )

21 11 21x

xx

xx

x

− − − − = =− + −

262. Ans. (b), ∵ ( ) ( )sin 120 sin 120α β° − = °− [∵ We know that ( )sin sin 180θ θ= °− ] ∴ 120 120α β° − = ° − or ( )120 180 120α β° − = °− °− ⇒ α β= or 60 3πα β+ = ° = 263. Ans. (b), Given, 3sin 2 5 4cos 2θ θ= + 2

2 22 tan 1 tan3 4 51 tan 1 tanθ θθ θ

− ⇒ − = + + 2

2 22 tan 1 tanWe know that sin 2 and cos 21 tan 1 tanθ θθ θ

θ θ −

= = + + ∵

2 26 tan 4 4 tan 5 5 tanθ θ θ⇒ − + = + 2tan 6 tan 9 0θ θ⇒ − + = ( )2tan 3 0 tan 3θ θ⇒ − = ⇒ = 264. Ans. (a), Given Expression ( ) ( )tan 160 110 tan 50= °− ° = ° ( )tan 90 40= °− ° cot 40= °



( ) 2

1 12 tan 20tan 2 20 1 tan 20= =

°× °− °

21 tan 202 tan 20− °=

° 2211 1212

pp

p

p

−−

= =

1cot 20 tan 20pp

= ° ⇒ ° =

∵

265. Ans. (b), Given, ( )( )

sinsin x y a b

x y a b

+ +=

− −

⇒ ( ) ( )( ) ( )

sin sinsin sinx y x y

x y x y

+ + −

+ − − ( ) ( )

( ) ( )a b a b

a b a b

+ + −=

+ − − [By Componendo and Dividendo]

⇒2sin cos 22cos sin 2x y a

x y b

⋅=

⋅ ⇒ tantan x a

y b=

266. Ans. (c), We note that ( )1sin15 cos15 2 sin15 cos152° ⋅ ° = ⋅ ° ⋅ ° ( )1 1 1sin 302 2 2 = ° =

1 ,4= which is a rational number, Hence option (c) is correct

Note that 3 1sin15 2 2−° = , clearly not a rational number 3 1cos15 2 2+° = , clearly not a rational number and 22 3 1sin15 cos 75 sin15 sin15 sin 15 2 2 −

°⋅ ° = °⋅ ° = ° =

, Hence, options (a), (b), (d) are wrong. 267. Ans. (b), Given Expression 1 cos 1 cos1 cos 1 cosα α

α α− +

= ++ −

( ) ( )2

1 cos 1 cos1 cosα α

α

− + +=

−

22 2sinsin αα

= = 2sinα=−

, Since sinα is negative in 3rd quadrant 2cosecα= − 268. Ans. (d), ∵ ( )cot 0 cot 90α β+ = = ° 90α β⇒ + = ° ∴ ( ) ( )sin 2 sinα β α β β+ = + + ( )sin 90 cosβ β= °+ = 269. Ans. (c), Given, cosec cot pθ θ− = …(i) As we know that 1cosec cot cosec cotθ θ

θ θ+ =

− 1cosec cot

pθ θ⇒ + = …(ii)

∴ Adding (i) and (ii) we get, 12cosec pp

θ = + 1 1cosec 2 pp

θ

⇒ = +

270. Ans. (d), Given, sec tan xθ θ− = …(i) As we know that 1sec tan sec tanθ θ

θ θ+ =

− ⇒ 1sec tan

xθ θ+ = …(ii)

Now, adding we get, 21 12sec xxx x

θ+

= + = ∴ 22cos 1xxθ =+

Then 2 2sin 1 cosθ θ= − ( ) ( )

( )( ) ( )

( )

2 22 2 222 22 2 2

1 2 1 2 1 221 1 1 1x x x x x xx

x x x

+ − + − + + = − = = + + +

22211 x

x

−= +

2

21sin 1 x

xθ

−⇒ =

+



271. Ans. (b), Given ( ) ( )cos105 sin105 cos 90 15 sin 90 15⇒ °+ ° = °+ ° + °+ ° 3 1 3 1 2 1sin15 cos15 cos 452 2 2 2 2 2 2 − +

= − °+ ° = − + = = = °

, Hence, option (b) is correct. Alternatively : ( )cos105 sin105 cos105 sin 90 15°+ ° = °+ °+ ° cos105 cos15= °− ° 105 15 105 152sin sin2 2° + ° ° − ° = ⋅

2sin 60 sin 45= °⋅ °

1 1 12 cos 452 2 2= × × = = ° Hence, option (b) is correct. 272. Ans. (c), Given Expression 2 58 2cos 16 2sin sin2 2 2A A A = − ⋅

( ) ( )8 1 cos 16 cos 2 cos3A A A= + − − ( ) ( )2 38 1 cos 16 2cos 1 16 4cos 3cosA A A A = + − − − − 2 33 3 3 38 1 16 2 1 16 4 34 4 4 4 = + − × − − − ×

( )14 18 16 27 36 3= − − − − =

273. Ans. (c), ∵ ( )sin 1α β+ = ∴ 2πα β+ = …(i) and ( ) 1sin 2α β− = ∴ 6πα β− = …(ii) Solving (i) and (ii) we get, 3πα = and 6πβ = ∴ ( ) ( )tan 2 tan 2α β α β+ ⋅ + 2tan tan3 3 3 6π π π π = + ⋅ +

2 5tan tan3 6π π =

tan tan3 6π ππ π = − ⋅ −

tan tan3 6π π = − −

( )1 3 13 = − − =

274. Ans. (c), Given, sin 25 cos 25 p° + ° = ( )2 2sin 25 cos 25 p⇒ °+ ° = [On squaring] ∴ ( ) ( )2 2 2sin 25 cos 25 2sin 25 cos 25 p° + ° + °⋅ ° = ( ) 21 sin 2 25 p⇒ + × ° = 2sin 50 1p⇒ ° = − cos50⇒ ° = 21 sin 50+ − ° ( )221 1p= − − ( )4 21 2 1p p= − − + 4 21 2 1p p= − + − 2 4 22 2p p p p= − = ⋅ − 275. Ans. (b), In the first quadrant, as the value of x goes from c0 0= ° to c90 2π ° =

the value of sin x

increase from 0 to 1. Now, 1 57c ≈ ° so that 1 1c° < and hence, sin1 sin1c° < 276. Ans. (b), Given 1 sin 1 sin 1 sin 1 sin1 sin 1 sin 1 sin 1 sinx x x x

x x x x

− + − += + = +

+ − + − ( ) ( )

( ) 221 sin 1 sin 2cos1 sinx x

xx

− + += =

−

2 2cos cosx x= =

− 2sec x= − [∵ cos 0x < in 2nd quadrant]

277. Ans. (b), We know that ( )sin cos cos sin sinx y x y x y⋅ − ⋅ = − We get, ( )sin sin3 3A B A B

π π + − + = −

278. Ans. (d), ∵ 0 2B

π< < 02 B

π⇒ − < − < also 0 2A

π< <

Hence, 2 4A Bπ π

− < − < A B⇒ − lies in 1st or 4th quadrant



279. Ans. (a), Given, A and B in the 1st quadrant, so all six ratios are positive, ∴ cos 0A > and sin 0B > ∴ ( )22cos 1 sin 1 3 / 5 4 / 5A A= + − = − = , Also ( )22sin 1 cos 1 9 / 41 40 / 41B B= + − = − = ∴ ( )sin sin cos cos sinA B A B A B− = ⋅ − ⋅ ( ) ( ) ( )( )3 / 5 9 / 41 4 / 5 40 / 41= − 133 / 205= − 280. Ans. (a), ( )1 tan tan / 2A A+ ⋅ ( )

( )sin / 2sin1 cos cos / 2AA

A A

= + ⋅

( ) ( )( )

cos cos / 2 sin sin / 2cos cos / 2A A A A

A A

⋅ + ⋅=

⋅ ( )

( )cos / 2 1 seccos cos / 2 cosA A

AA A A

−= = =

⋅

281. Ans. (a), ∵ ( )sin sin cos cos sincos cos cos cosA B A B A B

A B A B

− ⋅ − ⋅=

⋅ ⋅ sin sincos cosA B

A B= − tan tanA B= −

Similarly, simplify the other two terms. We get, ( )sin tan tancos cosB CB C

B C

−= − and

( )sin tan tancos cosC AC A

C A

−= −

⋅, Then, the given expression becomes

( ) ( ) ( )tan tan tan tan tan tanA B B C C A= − + − + − = 0 282. Ans. (d), ∵ ( ) ( )2sin 60 cos 30x x+ ° = − ° ( )2 sin cos60 cos sin 60x x⇒ ⋅ °+ ⋅ ° =

1 32 sin cos2 2x x

⋅ + ⋅

3 1cos sin2 2x x= ⋅ + ⋅ ∴ 1 3sin cos2 2x x⋅ = − ⋅ ∴ tan 3x = − 283. Ans. (c), ∵ ( ) ( )sin 2 sinθ φ θ φ+ = ⋅ − ∴ sin cos cos sinθ φ θ φ⋅ + ⋅ 2sin cos 2cos sinθ φ θ φ= ⋅ − ⋅ 3cos sin sin cosθ φ θ φ⇒ ⋅ = ⋅ 3 tan tanφ θ⇒ = tan 3tanθ φ⇒ = , So clearly, 3k = 284. Ans. (b), ∵ 3 tan tan 1θ φ⋅ = sin sin3 1cos cosθ φ

θ φ⋅

⇒ =⋅

cos cos 3sin sin 1θ φθ φ⋅

⇒ =⋅

By Componendo and dividendo, we get, cos cos sin sin 3 1cos cos sin sin 3 1θ φ θ φ

θ φ θ φ⋅ + ⋅ +

=⋅ − ⋅ −

( )( )

cos 2cos θ φθ φ−

⇒ =+

285. Ans. (b), We have ( )cot 54 cot 90 36 tan 36° = ° − ° = ° and ( )tan 20 tan 90 70 cot 70° = °− ° = ° ∴ Given expression cot 54 tan 20tan 36 cot 70° °

= +° °

tan 36 cot 70tan 36 cot 70° °= +

° ° 1 1 2= + =

286. Ans. (a), Use 4 2 4π π πθ θ + = − −

, we get, given expression 2 2cos sin 14 4π π

θ θ = − + − =

287. Ans. (c), 3cot cot cot cot20 20 4 20π π π π

π3 −

cot 2 20π π −

3 3cot cot tan tan 120 20 20 20π π π π= ⋅ × × =

288. Ans. (d), Given Expression ( ) ( )2 2 2 2sin 5 sin 85 sin 10 sin 80 ......= °+ ° + °+ ° + ( )2 2 2 2sin 40 sin 50 sin 45 sin 90+ °+ ° + ° + ° ( ) ( )2 2 2 2sin 5 cos 5 .... sin 40 cos 40 = ° + ° + + ° + ° ( )

2 21 12 + +

[ ] 11 1 1 ... 8 times 12= + + + + + + 3 19 18 92 2 2= + = = 289. Ans. (c), ∵ θ is in second quadrant ∴ cos 0θ < and sin 0θ >



∴ 2sin 1 cosθ θ= + − 25 121 169 13= − = ∴ sin 2 2sin cosθ θ θ= ⋅ 12 52 13 13− =

120169= − 290. Ans. (c), ∵ 270 360θ° < < ° ∴ 135 1802θ ° < < °

∴ ( )/ 2θ lies in the second quadrant ∴ ( )sin / 2 0θ > and ( )cos / 2 0θ < ∴ 1 coscos 2 2θ θ+ = −

1 3/ 7 52 7+= − = −

291. Ans. (b), 2 2sin sin sin sin3 3 3A A Aπ π π − ⋅ + = −

23 sin4 A= −

∴ Given Expression 234sin sin4A A = −

33sin 4sinA A= − sin 3A= 292. Ans. (d), Given Expression ( )cos 20 cos 40 cos80 cos 60= °⋅ ° ⋅ ° ⋅ ° ( )1 1cos 3 204 2= × ° ⋅ ( )1 cos 3 208= ⋅ × ° 41 18 2 16 21 1

= × = = 293. Ans. (c), We know that, tan tan tan tan 33 3A A A A

π π ⋅ − ⋅ + =

∴ tan 60 tan 20 tan 40 tan80⋅ ⋅ ° ⋅ ° ( ) ( )3 tan 20 tan 60 20 tan 60 20= °⋅ ° − ° ⋅ ° + ° ( )3 tan 3 20= × ° 3 3 3= ⋅ = 294. Ans. (c), ( )

( )22 2 32 tan 3tan 2 1 tan 41 3αα

α= = = −

− −

Now, dividing Nr. and Dr. of given expression by cos 2 ,α we get ( )

( )2 3/ 4 32 tan 2 34 tan 2 5 4 3/ 4 5α

α− −−

= =+ − +

6 12 1812 20 8− −= = −− +

94= − 295. Ans. (b), Given Expression 2 2 2cos cos cos8 2 8 2 8π π π π π = + − + +

2cos 8ππ + −

2 2 2 2cos sin sin cos8 8 8 8π π π π= + + + 2 22 cos sin8 8π π = +

( )2 1 2= =

296. Ans. (b), The given expression is ( ) ( ) ( )cos80 cos 40 cos 180 20 cos 180 60= °+ ° + °− ° + °+ ° 1 12cos 60 cos 20 cos 20 cos 60 2 cos 20 cos 202 2= °⋅ ° − ° − ° = × ° − ° − 1 1cos 20 cos 20 2 2= °− ° − = − 297. Ans. (c), ∵ ( )cos 40 cos 90 50 sin 50° = ° − ° = ° and ( )sin 40 sin 90 50 cos50° = ° − ° = ° ∴ Given expression sin 70 sin 50 2 sin 60 cos10cos 70 cos50 2 cos 60 cos10° + ° ⋅ ° ⋅ °

= =° + ° ⋅ ° ⋅ °

tan 60 3= ° = 298. Ans. (d), ∵ sin 5sin 3α

β= ∴ By Componendo and Dividendo sin sin 5 3sin sin 5 3α β

α β+ +

=− −

∴ 2sin cos 82 2 22cos sin2 2

α β α β

α β α β

+ − ⋅ =

+ − ⋅

∴ tan 2 4tan 2α β

α β

+ =

−

299. Ans. (b), ( )cos172 cos 180 8 cos8° = ° − ° = − °



∴ Given Expression ( )cos52 cos68 cos8= °+ °− ° cos52 2 sin 38 sin 30= °− ⋅ ° ⋅ ° ( ) 1cos52 2 sin 90 52 2= °− ⋅ ° − ° ⋅ cos52 cos52 0= °− ° = 300. Ans. (a), We have 4 2 2 2sin sin cos cosN α α α α= + ⋅ + ( )2 2 2 2sin sin cos cosα α α α= + + 2 2sin cosα α= + 1= Similarly, we can show that 1D = ∴ Given Expression 1/1 1= 301. Ans. (c), The given expression is ( ) ( )2 2 2 2 2 2sin cosec 2 cos sec 2 tan cotα α α α α α= + + + + + − − ( ) ( )2 2 2 24 sin cos sec tanα α α α= + + + − ( )2 2cosec cotα α+ − ( ) ( ) ( )4 1 1 1 7= + + + = 302. Ans. (b), ( )sin163 sin 180 17 sin17° = ° − ° = ° , ( )cos347 cos 2 180 13 cos13° = × °− ° = ° and ( )sin 73 sin 90 17 cos17° = ° − ° = ° , ( )sin167 sin 180 13 sin13° = ° − ° = ° ∴ Given Expression sin17 cos13 cos17 sin13= °⋅ ° + °⋅ ° ( )sin 17 13= °+ ° sin 30= ° 1 / 2= 303. Ans. (d), Note that 99 99 99sin sin sin 3a b c+ + = is possible iff sin sin sin 1,a b c= = = i.e., 90a b c= = = ° ∵ cos90 0° = ∴ 100 100 100cos cos cos 0 0 0 0a b c+ + = + + = 304. Ans. (c), ( ) ( )2 2sin sin sin sinN A B A B A B= − = + ⋅ − , sin cos sin cosD A A B B= ⋅ − ⋅ ( ) ( )1 2sin cos 2sin cos2 A A B B= ⋅ − ⋅ [ ]1 sin 2 sin 22 A B= − ( ) ( )1 2 cos sin2 A B A B= ⋅ + ⋅ − ( ) ( )cos sinA B A B= + ⋅ − ∴ ( ) ( )

( ) ( )sin sincos sinA B A BN

D A B A B

+ ⋅ −=

+ ⋅ − ( )tan A B= +

305. Ans. (c), Given, ( )sin 1A B C+ + = ⇒ 90A B C+ + = ° and ( ) 1tan 3A B− = 30A B⇒ − = ° , also ( )sec 2A C+ = 60A C⇒ + = ° ⇒

1cos ( ) 2A C+ = ∴ 30 , 60B A= ° = ° and 0C = ° 306. Ans. (c), Given, 90A = ° ∴ 90B C+ = ° 90C B⇒ = °− [ ]180A B C+ + = °∵ Operating cos on both sides we get, ( )cos cos 90 sinC B B= °− = ⇒ 2 2 2 2cos cos cos sin 1B C B B+ = + = 307. Ans. (c, d), Each one of the sequences, , andx y z is an infinite G.P. with first term 1 and common ratio 1.< 2 2 20

1 1cos 1 cos sinn

n

x θθ θ

∞

=

∴ = = =−∑ 2and 1, cos1 aS a r

rθ∞

= = = − ∵

2 2 201 1sin 1 sin cosn

n

y θθ θ

∞

=

= = =−∑ 2and 1, sin1 aS a r

rθ∞

= = = − ∵

2 2 2 201cos sin 1 cos sinn n

n

z θ θθ θ

∞

=

= =−∑ 2 2and 1, cos sin1 aS a r

rθ θ∞

= = = − ∵

Note that 2 2sin cos 1θ θ+ = 1 1 1x y

⇒ + = x y xy⇒ + = …(i) Now to get relation in , ,x y z put 2 1sin

xθ = and 2 1cos

yθ = in 2 211 sin cosz

θ θ=

− ⋅



We get, 11 11x

x y

=− ⋅

xyz xy z⇒ = + , Hence option (c) is true. Using (i) we get, ( )xyz x y z= + + , Hence option (d) is also correct. 308. Ans. (c), Given that diameter of circular wire 10= cm ∴Length of wire 10π= Hence, required angle arc 10radius 50 5π π

= = = rad 309. Ans. (c), we know that 2 2 2 2sin cosa b a b a bθ− + ≤ + ≤ + Maximum value of 2 2 23cos 4sin 3 4 5 5θ θ+ = + = = 310. Ans. (c), Given that, sin cos mθ θ+ = …….. (i) And sec cosec nθ θ+ = …. (ii) Now, ( ) ( ) ( )21 1 1n m m n m+ − = − ( )sec cosec 2sin cosθ θ θ θ= + ( )2 1 2sin cosm θ θ= +∵ sin cos 2sin cos 2sin cos m

θ θθ θ

θ θ+

= ⋅ = 311. Ans. (b), We have, ( ) ( )sin 420 cos390 cos 300 sin 330° ⋅ ° + − ° ⋅ − ° ( ) ( )sin 60 cos30 cos 360 60 sin 360 30= °⋅ °− ° − ° ⋅ ° − ° ( )sin 60 cos30 cos60 sin 30= °⋅ °− °⋅ − ° sin 60 cos30 cos 60 sin 30= °⋅ ° + °⋅ ° ( )sin 60 30 sin 90 1= °+ ° = ° = 312. Ans. (c), Now, ( ) ( )cot 45 cot 45θ θ° + ° − ( ) ( )tan 90 45 cot 45θ θ= °− °− °− ( ) ( )tan 45 cot 45 1θ θ= °− °− = 313. Ans. (c), Now, ( ) ( )sin sin 60 cot 60A A A° − °+ ( )2 2sin sin 60 sinA A= °− 23sin sin4A A

= −

33sin 4sin sin 34 4A A A−= =

314. Ans. (d), ( )f A sin cossin cos cos sincos sinA AA A A A

A A= ⋅ ⋅ + ⋅ ⋅ 2 2sin cos 1A A= + =

315. Ans. (c), Let ( ) ( )3cos 4sin 5f x x x= + + We know that, 2 2 2 2cos sina b a x b x a b− + ≤ + ≤ + 2 2 2 23 4 3cos 4sin 3 4x x⇒ − + ≤ + ≤ + 25 3cos 4sin 25x x⇒ − ≤ + ≤ 5 3cos 4sin 5x x⇒ − ≤ + ≤ ⇒ ( )5 5 3cos 4sin 5 5 5x x− + ≤ + + ≤ + ( )0 10f x⇒ ≤ ≤ ∴ The maximum value of ( )f x is 10 316. Ans. (b), sin 1 cos1 cos sinx x

x x

++

+ ( )

( )

22sin 1 cossin 1 cosx x

x x

+ +=

+ ( )1 1 2cossin 1 cos xx+ +

=+

( )( )

2 1 cossin 1 cosxx x

+=

+ 2sin x=

12 2 cosec sin xx

= = 317. Ans. (a), Let ( ) sin 3 cos 2 cos3 sin 2f θ θ θ θ θ= ⋅ + ⋅ ( )sin 3 2θ θ= + sin 5θ= ( )sin sin .cos cos sinA B A B A B+ = + ⋅ ∵ We know that 1 sin 5 1θ− ≤ ≤ ( )1 1f θ⇒ − ≤ ≤ So, the maximum value of ( )f θ is 1



318. Ans. (c), cos12 sin12 sin147cos12 sin12 cos147° − ° °+

° + ° °

1 tan12 tan1471 tan12− °= + °

+ °( ) ( )tan 45 12 tan 180 33= °− ° + °− ° tan 33 tan 33 0= °− ° =

319. Ans. (d), Let 12A∠ = rad, 99 1199 180 20Bπ π°×

∠ = ° = =°

We know that, A B C π∠ +∠ +∠ = 1 112 20 C

ππ⇒ + +∠ = 11 1 9 1020 2 20C

π ππ

−⇒ ∠ = − − =

320. Ans. (c), clearly 2 2sec cos 2, 0 2πθ θ θ+ ≥ ∀ < < 1we know that 2 if 0x xx

+ ≥ > ∵

321. Ans. (a), ∵ 3tan 4A = and 12tan 5B = − ∴ ( )

( )1 1 tan tancot tan tan tanA B

A BA B A B

+− = =

− −

Which shows that ( )cot A B− has only one value of A and B 322. Ans. (b), 1 13 3 1sin cos tan 2 24 4 4 13 3 3 2 2 1sec cosec cot4 4 4

π π π

π π π

3 − ++ −= =− + ++ −

323. Ans. (b), We know that, 1 1° < rad sin1 sin1⇒ ° < 324. Ans. (c), cos15 sin15 cos 45 cos15cos 45 sin 45 sin 45 sin15° ° ° °

×° ° ° °

( )sin 45 cos15 cos 45 sin15= ° °− ° ° × ( )cos 45 sin15 sin 45 cos15° ° − ° ° ( ) ( )sin 45 15 sin 15 45= °− ° × °− ° sin 30 sin 30= − °× ° 1 1 12 2 4= − × = − 325. Ans. (c), ∵ 24sin 4cos 1 0x x+ − = 24 4cos 4cos 1 0x x⇒ − + − = 24cos 4cos 3 0x x⇒ − + + = 24cos 4cos 3 0x x⇒ − − = 24cos 6cos 2cos 3 0x x x⇒ − + − = ( )( )2cos 3 2cos 1 0x x⇒ − + = 3cos 2x⇒ = (which is not possible) And 1cos 2x = − 1cos cos 2102⇒ = − = ° ( ) A lies in IIIrd quadrant∵ 210A⇒ = ° 326. Ans. (b), Given that, 1 1cos 2 x

xθ = +

2cosx

xθ

1⇒ + = …(i)

We know that, 22 21 1 2x xx x

+ = + −

( )2 22cos 2 4cos 2θ θ= − = − 2cos 2θ= [From eq. (i)]



∴ 2 21 1 1 2cos 2 cos 22 2xx

θ θ + = × =

327. Ans. (a), Let ( ) 3 cos sinf x x x= + ( ) 3 12 cos sin 2sin2 2 3f x x x x

π ⇒ = + = +

Since, 1 sin 13xπ − ≤ + ≤

Hence, ( )f x is maximum, if 3 2x

π π+ = 306x

π⇒ = = °

Or directly use 2 2 2 2sin cosa b a b a bθ θ− + ≤ + ≤ + Hence 1 3 sin 3 cos 1 3x x− + ≤ + ≤ + 328. Ans. (d), Given that, ABCD is a cyclic quadrilateral So, 180A C+ = ° 180A C⇒ = °− ( )cos cos 180 cosA C C⇒ = °− = − cos cos 0A C⇒ + = … (i) Similarly, cos cos 0B D+ = …. (ii) On adding eqs. (i) and (ii), we get cos cos cos cos 0A B C D+ + + = 329. Ans. (b), Clearly 2

31 3tan 13tan tan tan 3A

A A A

−=

−141tan 4π

= 4112Aπ =

∵

1 tan 4tan 10 4π

ππ

= = +

1= 330. Ans. (a), Arc length of circle 5 15 180r

πθ = × °×

° ( )15θ = °∵

512π= cm 331. Ans. (d), 1 sin10 sin 50 sin 70− ° ° ° ( ) ( )1 11 4sin10 sin 60 10 sin 60 10 1 sin 304 4= − ⋅ ° ° − ° ° + ° = − ° 1 71 8 8= − = 332. Ans. (a), ∵ 5sin 13θ = and 99sin 101φ = ∴ ( ){ }cos π θ φ− + ( )cos θ φ= − + { }cos cos sin sinθ φ θ φ= − − 225 99 5 991 1169 101 13 101 = − − − − ×

12 20 5 9913 101 13 101 = − × − ×

240 495 2551313 1313 1313 = − − =

333. Ans. (a), ∵ sin cosx θ θ= and sin cosy θ θ= + ∴ ( )22 2 sin cos 2sin cosy x θ θ θ θ− = + − 2 2sin cos 2sin cos 2sin cosθ θ θ θ θ θ= + + − 2 2sin cos 1θ θ= + = 334. Ans. (d), 4 4sin cosx x p− = (given) ( )( )2 2 2 2sin cos sin cosx x x x p⇒ − + = 2 2sin cosx x p⇒ − = cos 2x p⇒ − = cos 2x p⇒ = − 1 cos 2 1 cos 2 1x x − ≤ ≤ ⇒ ≤ ∵



∴ 1p ≤ 335. Ans. (d), Given that 1sin 10A = and 1sin 5B = We know that ( )sin sin cos cos sinA B A B A B+ = + 1 1 1 11 15 1010 5= − + − 1 4 9 15 1010 5= + ( )1 5 12 350 50 2⇒ + = = ( )sin sin 4A B

π⇒ + = 4A B

π⇒ + =

336. Ans. (b), We know that, 1 57 17c ′= ° ∴ ( )

sin1 sin1 1sin1 sin 5727c

° °= <

° [ ]sin1 sin 57 17′° < °∵

337. Ans. (a), The given expression is, ( ) ( )sin sin 120 sin 240θ θ θ+ + ° + + °

2 4sin sin sin3 3π πθ θ θ

= + + + +

( ) ( )sin 2sin cos sin sin3πθ θ π θ π θ= + + = + + sin sin 0θ θ= − = 338. Ans. (b), We know that 1 57 17c ′= ° (approximately) 2 114 34c ′⇒ = ° Hence correct choice is (b) 339. Ans. (c), ( ) ( ) ( ) ( )sin sin cos cosX Y A B A B A B A B+ = + − + + − ( ) ( ){ }cos A B A B= + − − cos 2X Y B⇒ + = (a) for 0° 45 ,B< < ° obviously 0X Y+ > (b) for 45 ,B = ° we see that 0X Y+ = (c) for 45 90 ,B° < ≤ ° clearly 0X Y+ < ∴ Only option (c) is incorrect for any A and ,B the expression dosen’t give any rational value for X Y+ 340. Ans. (a), 180tan tan tan1512 12π ° = = °

( ) tan 45 tan 30tan 45 30 1 tan 45 tan 30°− °= ° − ° =

+ °⋅ °

11 3 1 3 131 3 1 3 11 3−

− −= = ×

+ −+

( )23 1 3 1 2 3 2 33 1 2− + −= = = −

−

341. Ans. (a), ( )sin 2 sin 2π π≠ − ∵ ( )sin 2 sin 2π π− = − 342. Ans. (a), Given that, sec cos , sec sin , tans a y b z cθ φ θ φ θ= = = 2 2 2 2 2 2

2 2 2 2sec cosx y z a

a b c a

θ φ+ − =

2 2 2 2 22 2sec sin tanb c

b c

θ φ θ+ −

( )2 2 2 2sec cos sin tanθ φ φ θ= + − 2 2sec tan 1θ θ= − = 343. Ans. (c), Since, 2A B

π+ =

∴ cos cos cos cos2A B B Bπ = −

1sin cos sin 22B B B= =

∴ This greatest and the least values are 12 and 12− respectively



344. Ans. (d), Given tan tan , cot cotp qα β α β= + = + ∴ 1 1 1 1tan tan cot cotp q α β α β

− = − + +

1 tan tantan tan tan tanα βα β α β

= − + +

( )1 tan tan 1tan tan tanα β

α β α β−

= =+ +

( )cot α β= + 345. Ans. (b), ( ) ( ){ } ( )

( ) ( ) ( ){ }

22cosec cot 9 / 2 cosec 2cot 2 sec sec 3 / 2π θ π θ π θ

π θ π θ π θ

+ − −

− ⋅ − + 9cot cot tan2 2π π

θ θ θ − = − = ∵

( )( )( )( ) ( )( )2cosec tan coseccot sec cosecθ θ θ

θ θ θ

−=

− 2 2 2 2

2 2 2tan cosec sin cos 1sec cos sinθ θ θ θθ θ θ

⋅= = ⋅ =

346. Ans. (c), sin 20 sin 40 sin80° ° ° ( )1 sin 20 sin 60 2sin 40 sin 802= ° ° ° ° ( )1 sin 20 sin 60 cos 40 cos1202= ° ° ° − ° 21 3 1sin 20 1 2sin 202 2 2 = ⋅ ° − ° +

( )33 3sin 20 4sin 208= °− ° 3 3 3 3sin 608 8 2 16= ° = ⋅ = Or diretly use the paraula ( ) ( ) 1sin sin 60 sin 60 sin 34θ θ θ θ− ⋅ + = 347. Ans. (b), We have, ( )4 433 sin sin 32π α π α

− + + ( )6 62 sin sin 52π α π α

− + + −

( ) ( )4 4 6 63 cos sin 2 cos sinα α α α = − + − − + ( ) ( )32 2 2 2 2 22 cos sin 3cos sin cos sinα α α α α α − + − +

2 2 2 23 6sin cos 2 6sin cosα α α α= − − + 3 2 1= − = 348. Ans. (a), tan15 tan195° ⋅ ° ( )tan15 tan 180 15= °⋅ ° + ° tan15 tan15= °⋅ ° ( )2tan 15= ° …. (i) Clearly ( )( ) 22 tan15 1tan 2 15 tan 30 1 tan 15 3°

° = ° = =− °

22 tantan 2 1 tanθθθ

= − ∵

21 tan 15 2 3 tan15⇒ − ° = ° 2tan 15 2 3 tan15 1 0⇒ °+ °− =

( )2 3 12 4 2 3 4tan15 3 22 1 2− ± + − ±

⇒ ° = = = − ± ( )tan15 3 2⇒ ° = − + rejecting 3 2− − From Eq. (i), we get ( )2tan15 tan195 2 3° ⋅ ° = − 4 3 4 3= + − ( )7 4 3= − 349. Ans. (d), Let 2 2 2 .....x = + + + ∞ ∴ 2x x= + On squaring both sides, we get 2 2x x= + 2 2 0x x⇒ − − = ( ) ( )1 2 0x x⇒ + + = 2, 1x x⇒ = = − ∴ 2x = is correct ( ) is positivex∵ ∴ 2 cosecθ= 1sin 2θ⇒ = 350. Ans. (a), The squares of the tangents of the angles 30 , 45° ° and 60° are in GP



2 2 2tan 30 , tan 45 , tan 60⇒ ° ° ° are GP ( )2 21 , 1, 33

⇒

are in GP 1 , 1, 32= are in GP (∵ Condition for GP, 2b ac= is satisfied ) i.e., 2 11 32= × 1 1⇒ = 351. Ans. (d), I. ( ) ( )1 1sec tan sec1200 tan1200θ θ − −

+ = °+ ° ( ) ( )

1sec 6 120 tan 6 120π π−

= + ° + + ° ( ) 1sec120 tan120 −= °+ °

( ) 1cosec 30° cot 30 −= − − ° ( ) 1 12 3 ,2 3−

= − − = −+

negative II. ( ) ( )cosec cot cosec 6 120 cot 6 120θ θ π π− = + ° − + ° cosec120° cot120= − ° sec30 tan 30= °+ ° 2 3,3= + positive Hence, both the statements are incorrect, 352. Ans. (b), A. ( ) tan 45 tan 30tan15 tan 45 30 1 tan 45 tan 30° − °

° = ° − ° =+ °⋅ °

11 3 1 3 131 3 1 3 11 3

−− −

= = ×+ −+

3 1 2 3 2 32+ −= = −

B. ( )tan 75 tan 45 30° = ° + ° tan 45 tan 301 tan 45 tan 30°+ °=

− °⋅ °

11 3 1 3 13 2 31 3 1 3 11 3+

+ += = × = +

− +−

C. ( ) ( ) tan 60 tan 45tan 105 tan 60 45 1 tan 60 tan 45° + °° = °+ ° =

− °⋅ °3 1 1 31 3 1 3+ +

= ×− +

4 2 3 2 32+= = − −

−

353. Ans. (b), Given, 18θ = ° Now, we have ( ) ( )2 24sin 2sin 4sin 18 2sin 18θ θ+ = ° + ° 5 1sin18 4 −

° = ∵

25 1 5 14 24 4 − − = +

( ) ( )5 1 2 5 5 1 3 5 5 14 116 2 2 2+ − − − −= ⋅ + = + =

354. Ans. (a), I. 3 1sin sin1512 2 2π −= ° =

∵ ( )sin15 sin 45 30° = °− ° sin 45 cos30 cos 45 sin 30= °⋅ ° − °⋅ °3 1 3 12 2 2 2 2 2−= − =

II 3 1cos cos1512 2 2π += ° =

∵ ( )cos15 cos 45 30° = ° − ° cos 45 cos30 sin 45 sin 30= °⋅ ° + °⋅ °3 1 3 12 2 2 2 2 2+= + =

III. cot cot15 2 312π = ° = +



∵ ( )cos15 tan 75 tan 45 30° = ° = ° + °

11tan 45 tan 30 311 tan 45 tan 30 1 3+

°+ °= =

− °⋅ ° −

3 1 2 33 1+= = +−

∴ The correct sequence is III > II > I 355. Ans. (a), Ist, 1 180π° = radian 3.14 0.017180= = radian IInd , we know that I radian 57 17′° Which is greater than 45° 356. Ans. (a), 4sin 5A = 16 3cos 1 25 5A⇒ = − − = − 12cos 13B = − 144 5sin 1 169 13B⇒ = + − = ∴ ( )sin sin cos cos sinA B A B A B+ = +

4 12 3 55 13 5 13 = × − − ×

48 15 6365 65− −= = −

357. Ans. (b,d) , We have, 322 4cπ π< < and 52 7 2c π

π < < 2c⇒ is in second quadrant and 7c is in first quadrant cos 2 0ca⇒ = < and sin 7 0cb = > 0ab⇒ < 358. Ans. (c) We have, ( )2

21 sin1 sin1 sin 1 sinθθθ θ

−−=

+ − 1 sin 1 sin1 sin cosθ θ

θ θ− −

⇒ =+

Hence 1 sin 1 sin1 sin cosθ θ

θ θ− −

=+ −

3 cos 02 2π πθ θ < < ⇒ <

∵ 1 sin sec tan1 sinθ θ θ

θ−

⇒ = − ++

359. Ans. (b) , We have, 1 1 sin ,1 1 siny

y

θθ

+ +=

− −

22

cos sin1 2 21 cos sin2 2y

y

θ θ

θ θ

+ + ⇒ =− −


y

θ θ

θ θ

++⇒ =

− −


y

θ θ

θ θ

++⇒ =

− − 0 02 2 4cos sin2 2

π π πθ

θ θ

< < ⇒ < < ⇒ >

∵

1 tan1 21 1 tan 2y

y

θ

θ

++⇒ =

− − tan 2y

θ⇒ =

360. Ans. (d) Clearly, sec tanα α− is not defined for / 2.α π= ± Now, ( )2

21 sin1 sin1 sin 1 sinααα α

−−=

+ + 1 sin 1 sin sec tan ,1 sin cosα α

α αα α

− −⇒ = = −

+ if cos 0α >



Clearly, cos 0α > ( )/ 2, / 2α π π⇒ ∈ − 361. Ans. (a)We have, ( )2 2 1sin cos cos sin sin 44a b x x x x x− = − = 0a b⇒ − > for 0 4x

π< <

Also, 1sin cos sin 22a b x x x+ = = 0a b⇒ + > for 0 2x π< < i.e. 0 2xπ

< < 362. Ans. (b) We have, 3 3

2sin cos sin 2 tan cot .sin cos 1 tanA A AA A

A A A

−+ −

− +

( )2 2 sinsin cos sin cos 2sec AA A A AA

= + + + − 1 sin cos sin cos 2 1.A A A A= + − − = − 363. Ans. (b) We have, tan 0θ < θ⇒ lies either in II quadrant of IV quadrant sin 0θ⇒ > or, sin 0θ < 4tan 5θ∴ = − 4sin 5θ⇒ = ± 4sin 5θ⇒ = or, 4sin 5θ = − 364. Ans. (b) We have, ( )

( )sinsin x y a b

x y a b

+ +=

− −

( ) ( )( ) ( )

( ) ( )( ) ( )

sin sinsin sinx y x y a b a b

x y x y a b a b

+ + − + + −⇒ =

+ − − + − − 2sin cos 22cos sin 2x x a

x y b⇒ = tantan x a

y b⇒ =

365. Ans. (d), Given tan cos sin2x ec x x= − 2

21 tan 2 tan2 2tan 2 2 tan 1 tan2 2

x xx

x x

+⇒ = −

+ ( ) ( )22 1 1 ,t t t⇒ + = − where 2tan 2xt = 2 4 1 0t t⇒ + − =

2 5t⇒ = − + 2tan 02xt = > ∵

366. Ans. (c) We have, 532sin sin2 2A A ( )16 cos 2 cos3A A= − ( )2 316 2cos 1 4cos 3cosA A A= − − + 9 27 316 2 1 4 3 1112 64 4 = × − − × + × =

367. Ans. (a) We have, sin 2 cos3θ θ= 32sin cos 4cos 3cosθ θ θ θ⇒ = − ( )22sin 4 1 sin 3θ θ⇒ = − − 24sin 2sin 1 0θ θ⇒ + − = 5 1sin 4θ

−⇒ = Rejecting 5 14− − ∵θ is acute

368. Ans. (d) We have, 2 4 8 14cos cos cos cos15 15 15 15π π π



2 4 8cos cos cos cos15 15 15 15π π π π= − 4

416 16sin 2 sin sin 115 15 15 162 sin 16sin 16sin15 15 15

π π π

π π π

⋅ = − = − = − =

369. Ans. (b) We have, 2 3 4 5 6 7cos cos cos cos cos cos cos15 15 15 15 15 15 15π π π π π π π 2 4 7 3 6 5cos cos cos cos cos cos cos15 15 15 15 15 15 15π π π π π π π = × ×

2 4 8 3 6cos cos cos cos cos cos cos15 15 15 15 8 15 3π π π π π π ππ

= − ×

2 4 8 3 6 1cos cos cos cos cos cos15 15 15 15 15 15 2π π π π π π = − × ×

4 24 2

3sin 2 sin 2 115 153 22 sin 2 sin15 15π π

π π

× × = − × × 7

1216 sinsin 1 1 1 1 11515 3 2 16 4 2 216sin 4sin15 15ππ

π π

= − × × = × × =

370. Ans. (a) We know that ( ) 1 3 5 71 2 2 4 6....tan ..... 1 ....n

S S S S

S S Sθ θ θ

− + − ++ + + =

− + − +

5 5 3 5 51 3 55 2 52 4tan tan tantan 5 1 tan tanC C C

C C θ

θ θ θθ

θ− +

∴ =− +

371. Ans. (c) We have, 4 2 24sin sin 2 4cos 4 2π α

α α + + −

( )2 2 2 1 cos 24sin sin cos 4 2

πα

α α α

+ − = + +

( )2 sin 2 1 sinα α= + + ( )2sin 2 1 sin 2.α α= − + + = 3sin 0for , 2πα α π

< ∈

∵

372. Ans. (c) We have, 22sin cossin 2 2tan cos 1 2sin 2

α αα

ααα

= =−

2

21 12 12 2 111 2 2

x x

x xx

x

x

− − − = = −− −

373. Ans. (c) We have, sin cosA A m+ = and 3 3sin cos .A A n+ = Now, sin cosA A m+ = ( )3 3sin cosA A m⇒ + = ( )3 3 3sin cos 3sin cos sin cosA A A A A A m⇒ + + + = 33sin cosn A Am m⇒ + = …(i) Again, sin cosA A m+ = 2 2 2sin cos 2sin cosA A A A m⇒ + + = 2 1sin cos 2mA A

−⇒ = …(ii)



from (i) and (ii), we have ( )2 313 2mn m m−

+ = 3 32 3 3 2n m m m⇒ + − = 3 3 2 0m m n⇒ − + = 374. Ans. (d) cos 2 cos 2 cos 2A B C+ + ( ) ( )2cos cos cos 2A B A B C= + − + ( )32cos cos cos 22 C A B C

π = − − +

( ) 22sin cos 1 2sinC A B C= − − + − ( ){ }1 2sin cos sinC A B C= − − + ( ) ( )( ){ }1 2sin cos sin 3 / 2C A B A Bπ= − − + − + ( ) ( ){ }1 2sin cos cosC A B A B= − − − + 1 4sin sin sin .A B C= − 375. Ans. (a) We have, sin cos 2 cos6 6 6 4y x x x

π π π π = + + + = + −

2 cos 12xy x

⇒ = −

y⇒ is maximum for 012xπ

− = i.e. 12xπ

= 376. Ans. (d) We have, ( )2

21 cos1 cos1 cos 1 cosθθθ θ

++=

− − 1 cos 1 cos1 cos sinθ θ

θ θ+ +

⇒ =−

1 cos 1 cos1 cos sinθ θ

θ θ+ +

⇒ =− −

[ ]2 sin 0π θ π θ< < ⇒ <∵ 1 cos cos cot1 cos ec

θθ θ

θ+

⇒ = − −−

377. Ans.(b)We have, 1 sin 1 sin1 sin 1 sinθ θ

θ θ− +

++ −

21 sin 1 sin1 sinθ θ

θ

− + +=

−

2 2 2seccos cos θθ θ

= = = −−

/ 2cos 0π θ πθ< <

∴ <

∵ 378. Ans. (a) We have, ( ) ( ){ } 1tan tan cos sec 1 1cot cotα β

α β α βα β

−++ − + + =

+

( )( )

( )( ) ( )

sin cossin sin 1cos cos sin cos cosα β α βα βα β α β α β α β

+ +⇒ × + =

+ − + +

cos cos sin sintan tan 12cos cosα β α βα β

α β−

⇒ + = 1 1tan tan 12 2α β⇒ + = tan tan 1α β⇒ = 379. Ans. (a) We have, tan 70 tan 20tan 50° − °

° ( )sin 70 20 cos50cos 70 cos 20 sin 50° − ° °= ×

° ° ° 2cos50 2cos50 22cos70 cos 20 cos90 cos50° °= = =

° ° ° + °

380. Ans. (a) We have, 5 3sin 4cos 5θ θ− ≤ − ≤ for all θ 2 3sin 4cos 7 12θ θ⇒ ≤ − + ≤ for all θ 1 1 112 3sin 4cos 7 2θ θ

⇒ ≤ ≤− +

for all θ



381. Ans. (c) We have, cos 1 sincos 1 sin cosx xx

x x

− + −

( )( )

22cos 1 sincos 1 sin cosx xx

x x

+ − =

− 2 2sin 21 sin x

x

−= =

− for all x R∈ Hence, required value 2.=

382. Ans. (b) We have, tan bxa

= a b a b

a b a b

+ −∴ +

− + 1 1

1 1b b

a ab b

a a

+ −= +

− + 1 tan 1 tan1 tan 1 tanx x

x x

+ −= +

− + cos sin cos sincos sin cos sinx x x x

x x x x

+ −= +

− +

2 2cos sin cos sincos sinx x x x

x x

+ + −=

− 2 2

cos sin cos sin 2coscos 2cos sinx x x x x

xx x

+ + −= =

−

383. Ans. (c) We have, 1 sin 2sec 2 tan 2 cos 2 xx x

x

−− =

1 cos 22sec 2 tan 2 sin 22x

x x

x

π

π

− − ⇒ − =

−

( )( ) ( )

22sin / 4sec 2 tan 2 tan2sin / 4 cos / 4 4xx x x

x x

π ππ π

− ⇒ − = = − − −

384. Ans. (a) , We have, cos sina b

θ θ= tan b

aθ⇒ =

cos 2 sin 2sec 2 cos 2a ba b

b ecθ θ

θ∴ + = + 2

2 21 tan 2 tan1 tan 1 tana bθ θθ θ

−= +

+ +

( )2 2 22 2 2 22a b ab

a aa b a b

−= + =

+ +

385. Ans. (a) 2 2 2sin sin sinα β γ+ − ( ) ( )2sin sin sinα β γ β γ= + − + ( ) ( )2sin sin sinα π α β γ= + − + [ ]α β γ π+ − =∵ ( ){ }sin sin sinα α β γ= + + ( ) ( ){ }sin sin sinα β γ β γ= − + + ( )α π β γ= − − ∵ 2sin sin cosα β γ= 386. Ans. (d) We have, tan cot4 4απ βπ =

tan tan4 2 4απ π βπ ⇒ = −

4 2 4nπ π π

α π β ⇒ = + −

( )2 2 1nα β⇒ = + − ( )2 2 1nα β⇒ + = + 387. Ans. (a) We haves, coscos A n

B= and sinsin A m

B=

( ) ( ) ( ) ( )2 2 2 2sin sincos sinA B A Bm n m n m n

B B

+ −∴ − = + − = 2 22 2 2 2sin sincos sinA B

m nB B

−⇒ − =

( )2 2 2 22 2 2 2 2sin sin cos cossin cos cosA B B A

m n BB B

− −⇒ − = = ( )

22 2 2 22cossin 1 1cos Am n B nB

⇒ − = − = −



388. Ans.(c) We have, ( ) ( )cos cosmθ φ θ φ+ = − ( )( )

cos1 cosm

θ φθ φ−

⇒ =+

1 2cos cos1 2sin sinm

m

θ φθ φ

+⇒ =

− 1tan tan 1 m

mθ φ

−⇒ =

+ 1tan cot1 m

mθ φ

−⇒ =

+

389. Ans. (a), As 2 2cos sin 1x x+ = Hence 2 2 2 2cos sin 1a y b y+ = ∴ ( )2 2 2 21 sin sin 1a y b y− + = ∴ ( )2 2 2 2sin 1a a b y− − = ∴ ( )2 2 2 2sin 1a b y a− = − 390. Ans. (d), ( )2 2 1 cos 4 2 2 cos 2θ θ+ + = + ( )2 1 cos 2θ= − 22 2sin θ= ⋅ 2 sin 2sinθ θ= = as 32 4π π

θ< < Hence, (d) is the correct answer. 391. Ans. (c), 3 333cos cos3 4cos cot2sin sin 3 4sinθ θ θ

θθ θ θ+

= =−

392. Ans. (d), 2 2 2 23 4cos cos cos 108 cos 1445 5π π

+ = °+ ° 2 22 2 5 1 5 1sin 18 cos 36 4 4 − +

= °+ ° = +

( )2 5 1 316 4+= =

393. Ans. (d) 3sin 5sinα β= sin 5sin 3αβ

⇒ = sin sin 8sin sin 2α βα β+

⇒ =−

tan 2 4tan 2α β

α β

+ ⇒ =

−

394. Ans. (b),Given ( )sin 1A B+ = 2A B

π⇒ + = and ( ) 1sin 2A B− = 6A B

π⇒ − =

Hence we have 3Aπ

= and 6Bπ

= 395. Ans. (c), Now, ( ) ( )tan 2 tan 2A B A B+ ⋅ + 2tan tan3 3 3 6π π π π = + ⋅ +

2 5tan tan3 6π π = ⋅

tan tan2 6 2 3π π π π = + ⋅ +

( ) 1cot cot 3 16 3 3π π = − − = − ⋅ − =

396. Ans. (b), Now, 2 2sin sinA B− ( ) ( )2 2sin / 3 sin / 6π π= − 2 23 12 2 = −

3 1 2 14 4 4 2= − = =

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