BOARD OF INTERMEDIATE EDUCATION, KARACHI

INTERMEDIATE EXAMINATION, 2018 (ANNUAL) (Science pre-Engineering and Science General Groups)

XI-Mathematics

SOLUTION

FROM THE DESK OF: FAIZAN AHMED

Code: MT-09 Time: 20 Minutes

SECTION ‘A’ (MCQs-Multiple Choice Questions) (20 Marks)

Q1. Choose the correct answer for each from the given options.

i. If 1,x-1, 3 are in A.P., then 𝑥 =:

a. 0 b. 1 c. 2 d. 3 ii. The H.M.between 3 and 6 is:

a. 1

4 b.

9

2 c. ± 18 d. 4

iii. If 𝑎−𝑏

𝑏−𝑐=

𝑎

𝑏, then a,b,c are in:

a. A.P. b. 𝐺. 𝑃 . c. H.P. d. A.G.P iv. The number of permutations of the letters of the word COMMITTEE

a. 9

2,2,2 b.

62,2,2

c. 9

2,2,1 d.

2,2,29

v. The middle term in the expansion of 2𝑥 −1

𝑥2 20

is:

a. Ninth term b. tenth term c. 𝑒𝑙𝑒𝑣𝑒𝑛𝑡𝑕 𝑡𝑒𝑟𝑚 d. twelvth term

vi. If 𝑛 = 0, then 𝑛+1 !

𝑛 !=:

a. 0 b. 1 c. n d. ∞ vii. 𝑠𝑖𝑛600𝑐𝑜𝑠300 − 𝑐𝑜𝑠600𝑠𝑖𝑛300 =:

a. 1

2 b. −

3

2 c.

3

2 d. −

1

2

viii. If arc length s is equal to the radius r , then the central angle 𝜃 is:

a. 0 radian b. 1

2 radian c. 2 radian d. 1 𝑟𝑎𝑑𝑖𝑎𝑛

ix. In a triangle ABC, if 𝛾 = 900, then the law of cosine reduces to:

a. 𝑎2 = 𝑏2 + 𝑐2 b. 𝑏2 = 𝑎2 − 𝑐2 c. 𝑐2 = 𝑎2 + 𝑏2 d. 𝑐2 = 𝑎2 − 𝑏2

x. In an escribed triangle ABC, ∆

𝑟3=:

a. s b. (𝑠 − 𝑎) c. (𝑠 − 𝑏) d. (𝑠 − 𝑐)

xi. If 𝑟𝑐𝑜𝑠𝜃 = 4 and 𝑟𝑠𝑖𝑛𝜃 = 3, then 𝑟 =:

a. 3 b. 5 c. 6 d. 2 xii. 10.5 0 =:

a. 𝜋

18radians b.

7𝜋

120 radians c.

10.5

𝜋 radians d. 5𝜋 radians

xiii. If 𝐴 = {2,3} and 𝐵 = {3,4}, then 𝐴 − 𝐵 ∩ 𝐵 =:

a. ∅ b. {∅} c. {2} d. {3}

xiv. 𝐴𝑈𝐴′ ′ =:

a. 𝐴 b. 𝐴′ c. ∅ d. 𝑈

xv. The imaginary part of 𝑖 3 + 5𝑖2 is:

a. −2𝑖 b. 3𝑖 c. −2 d. −5 xvi. If 𝑧 is a complex number, then 𝑧𝑧 =:

a. 𝑧2 b. 𝑧 2 c. |𝑧| d. 𝑧 2

xvii. The product of the roots of the equation 𝑦2 + 1 = 7𝑦 − 7

a. 4 b. 8 c. 7 d. 1 xviii. If 𝜔 is a complex cube root of unity, then 2 − 𝜔 − 𝜔2 2 =:

a. −1 b. 1 c. 3 d. 9 xix. If A,B,C are non-singular matrices, then 𝐶𝐵𝐴 −1

a. 𝐴−1𝐵−1𝐶−1 b. 𝐶−1𝐵−1𝐴−1 c. 𝐴𝐵𝐶 −1 d. ABC xx. 𝐴 𝐴−1 =:

a. 𝐴𝐴−1 b. 𝐴 𝐼3 c. 𝑎𝑑𝑗𝐴 d. 𝐴2

SECTION B

COMPLEX NUMBER, ALGEBRA, MATRICES

Q.2(i) QUESTION: Solve the complex equation for x and y: 𝑥 + 2𝑦𝑖 2 = 𝑥𝑖. SOLUTION: (𝑥 + 2𝑦𝑖)2 = 𝑥𝑖

(𝑥)2 + 2 𝑥 2𝑦𝑖 + (2𝑦𝑖)2 = 𝑥𝑖

𝑥2 + 4𝑥𝑦𝑖 + 4𝑦2𝑖2 = 𝑥𝑖

𝑥2 + 4𝑥𝑦𝑖 + 4𝑦2(−1) = 𝑥𝑖 ∵ 𝑖 = −1 ⇒ 𝑖2 = −1

𝑥2 + 4𝑥𝑦𝑖 − 4𝑦2 = 𝑥𝑖

𝑥2 − 4𝑦2 + 4𝑥𝑦𝑖 = 0 + 𝑥𝑖

By comparing

4𝑥𝑦 = 𝑥 ⇒ 4𝑦 = 1

𝑦 =1

4

Now, 𝑥2 − 4𝑦2 = 0

𝑥2 − 4 1

4

2

= 0

𝑥2 −1

4= 0

𝑥2 =1

4

𝑥 = ±1

2

𝑆. 𝑆 = ±1

2,

1

4 OR 𝑆. 𝑆 =

1

2,

1

4 , −

1

2,

1

4

OR Q.2(i)

QUESTION: Solve the complex equation for x and y: 𝑥 1 + 2𝑖 + 𝑦 3 + 5𝑖 = −3𝑖 SOLUTION: 𝑥 1 + 2𝑖 + 𝑦 3 + 5𝑖 = −3𝑖 𝑥 + 2𝑥𝑖 + 3𝑦 + 5𝑦𝑖 = −3𝑖 𝑥 + 3𝑦 + 5𝑦𝑖 + 2𝑥𝑖 = −3𝑖 𝑥 + 3𝑦 + 𝑖 5𝑦 + 2𝑥 = 0 − 3𝑖 By Comparing 𝑥 + 3𝑦 = 0 𝑥 = −3𝑦 ---- 1 And 5𝑦 + 2𝑥 = −3 5𝑦 + 2(−3𝑦) = −3 5𝑦 − 6𝑦 = −3 −𝑦 = −3

𝑦 = 3

1 ⇒ 𝑥 = −3𝑦 𝑥 = −3 3 = −9 𝑆. 𝑆 = −9,3

2(ii) QUESTION:

Solve: 𝑥 +1

𝑥

2

= 4 𝑥 −1

𝑥

SOLUTION:

𝑥 +1

𝑥

2

= 4 𝑥 −1

𝑥

𝑥 2 + 2 𝑥 1

𝑥 +

1

𝑥

2

= 4 𝑥 −1

𝑥

𝑥2 + 2 +1

𝑥2= 4 𝑥 −

1

𝑥

𝑥2 − 2 +1

𝑥2+ 4 = 4 𝑥 −

1

𝑥

𝑥 −1

𝑥

2

+ 4 = 4 𝑥 −1

𝑥 ----(1)

Let 𝑦 = 𝑥 −1

𝑥

1 ⇒ 𝑦2 + 4 = 4𝑦 𝑦2 − 4𝑦 + 4 = 0 𝑦 − 2 2 = 0 𝑦 − 2 = 0

𝑦 = 2

But 𝑥 −1

𝑥= 𝑦

𝑥 −1

𝑥= 2

𝑥2 − 1

𝑥= 2

𝑥2 − 1 = 2𝑥 𝑥2 − 2𝑥 − 1 = 0

𝑥 =−𝑏± 𝑏2−4𝑎𝑐

2𝑎

𝑥 =−(−2)± (−2)2−4 1 (−1)

2(1)

𝑥 =2± 4+4

2=

2± 8

2=

2± 4×2

2=

2±2 2

2=

2 1± 2

2

𝑥 = 1 ± 2

𝑆. 𝑆 = 1 ± 2

2(iii) QUESTION: For what values of ‘m’ will the equation have equal roots? 𝑚 + 1 𝑥2 + 2 𝑚 + 3 𝑥 + 2𝑚 + 3 = 0 SOLUTION: 𝑚 + 1 𝑥2 + 2 𝑚 + 3 𝑥 + 2𝑚 + 3 = 0 ----(1)

𝐴 = 𝑚 + 1, 𝐵 = 2 𝑚 + 3 , 𝐶 = 2𝑚 + 3

As roots of (1) are equal

i.e. 𝐵2 − 4𝐴𝐶 = 0

2 𝑚 + 3 2 − 4 𝑚 + 1 2𝑚 + 3 = 0

4 𝑚2 + 6𝑚 + 9 − 4 2𝑚2 + 3𝑚 + 2𝑚 + 3 = 0

÷ 𝑖𝑛𝑔 𝑏𝑦 4, 𝑤𝑒 𝑔𝑒𝑡

𝑚2 + 6𝑚 + 9 − 2𝑚2 + 5𝑚 + 3 = 0

𝑚2 + 6𝑚 + 9 − 2𝑚2 − 5𝑚 − 3 = 0

−𝑚2 + 𝑚 + 6 = 0 ⟹ 𝑚2 − 𝑚 − 6 = 0

𝑚2 − 3𝑚 + 2𝑚 − 6 = 0

𝑚 𝑚 − 3 + 2 𝑚 − 3 = 0 𝑚 − 3 𝑚 + 2 = 0

𝐸𝑖𝑡𝑕𝑒𝑟: 𝑚 − 3 = 0 ⇒ 𝑚 = 3

𝑂𝑟: 𝑚 + 2 = 0 ⇒ 𝑚 = −2

2(iv) QUESTION:

If 𝑨 = 𝒔𝒊𝒏𝜽 −𝒄𝒐𝒔𝜽𝒄𝒐𝒔𝜽 𝒔𝒊𝒏𝜽

and 𝑩 = 𝒔𝒊𝒏𝜽 𝒄𝒐𝒔𝜽

−𝒄𝒐𝒔𝜽 𝒔𝒊𝒏𝜽 then prove that 𝑨𝑩 = 𝑩𝑨 = 𝑰𝟐

SOLUTION:

𝐴 = 𝑠𝑖𝑛𝜃 −𝑐𝑜𝑠𝜃𝑐𝑜𝑠𝜃 𝑠𝑖𝑛𝜃

and 𝐵 = 𝑠𝑖𝑛𝜃 𝑐𝑜𝑠𝜃

−𝑐𝑜𝑠𝜃 𝑠𝑖𝑛𝜃

??𝐴𝐵 = 𝐵𝐴 = 𝐼2

AB = 𝑠𝑖𝑛𝜃 −𝑐𝑜𝑠𝜃𝑐𝑜𝑠𝜃 𝑠𝑖𝑛𝜃

𝑠𝑖𝑛𝜃 𝑐𝑜𝑠𝜃

−𝑐𝑜𝑠𝜃 𝑠𝑖𝑛𝜃

= 𝑠𝑖𝑛2𝜃 + 𝑐𝑜𝑠2𝜃 𝑠𝑖𝑛𝜃𝑐𝑜𝑠𝜃 − 𝑠𝑖𝑛𝜃𝑐𝑜𝑠𝜃𝑠𝑖𝑛𝜃𝑐𝑜𝑠𝜃 − 𝑠𝑖𝑛𝜃𝑐𝑜𝑠𝜃 𝑐𝑜𝑠2𝜃 + 𝑠𝑖𝑛2𝜃

= 1 00 1

= 𝐼2

BA= 𝑠𝑖𝑛𝜃 𝑐𝑜𝑠𝜃

−𝑐𝑜𝑠𝜃 𝑠𝑖𝑛𝜃

𝑠𝑖𝑛𝜃 −𝑐𝑜𝑠𝜃𝑐𝑜𝑠𝜃 𝑠𝑖𝑛𝜃

= 𝑠𝑖𝑛2𝜃 + 𝑐𝑜𝑠2𝜃 −𝑠𝑖𝑛𝜃𝑐𝑜𝑠𝜃 + 𝑠𝑖𝑛𝜃𝑐𝑜𝑠𝜃−𝑠𝑖𝑛𝜃𝑐𝑜𝑠𝜃 + 𝑠𝑖𝑛𝜃𝑐𝑜𝑠𝜃 𝑐𝑜𝑠2𝜃 + 𝑠𝑖𝑛2𝜃

= 1 00 1

= 𝐼2

𝐴𝐵 = 𝐵𝐴 = 𝐼2

2(v) QUESTION: Using the properties of determinants, evaluate the determinant:

1 𝑥 𝑦𝑧1 𝑦 𝑧𝑥1 𝑧 𝑥𝑦

=

1 1 1𝑥 𝑦 𝑧

𝑥2 𝑦2 𝑧2

SOLUTION:

??

1 𝑥 𝑦𝑧1 𝑦 𝑧𝑥1 𝑧 𝑥𝑦

=

1 1 1𝑥 𝑦 𝑧

𝑥2 𝑦2 𝑧2

L.H.S=

1 𝑥 𝑦𝑧1 𝑦 𝑧𝑥1 𝑧 𝑥𝑦

× 𝑖𝑛𝑔 𝑅1 𝑏𝑦 𝑥, 𝑅2 𝑏𝑦 𝑦 𝑎𝑛𝑑 𝑅3 𝑏𝑦 𝑧

=1

𝑥𝑦𝑧

𝑥 𝑥2 𝑥𝑦𝑧

𝑦 𝑦2 𝑥𝑦𝑧

𝑧 𝑧2 𝑥𝑦𝑧

Taking common 𝑥𝑦𝑧 from 𝐶3

=𝑥𝑦𝑧

𝑥𝑦𝑧 𝑥 𝑥2 1𝑦 𝑦2 1

𝑧 𝑧2 1

= 𝑥 𝑥2 1𝑦 𝑦2 1

𝑧 𝑧2 1

Taking transpose

=

𝑥 𝑦 𝑧

𝑥2 𝑦2 𝑧2

1 1 1

Interchanging 𝑅2 and 𝑅3

= (−1)

𝑥 𝑦 𝑧1 1 1𝑥2 𝑦2 𝑧2

Interchanging 𝑅1 and 𝑅2

= (−1)(−1)

1 1 1𝑥 𝑦 𝑧

𝑥2 𝑦2 𝑧2

=

1 1 1𝑥 𝑦 𝑧

𝑥2 𝑦2 𝑧2 = 𝑅. 𝐻. 𝑆

SECTION B

GROUPS, SEQUENCES & SERIES, COUNTING PROBLEMS

Q.3(i) QUESTION: Let 𝐺 = {1, 𝜔, 𝜔2}, where 𝜔 is a complex cube root of unity. Show that 𝐺, . is an abelian group, where ‘.’ Is an ordinary multiplication. Note: For a group:

a) Associative law holds b) There exist an identity element w.r.t multiplication c) Every element has an inverse in G

For Abelian: multiplication is Commutative SOLUTION: 𝐺 = {1, 𝜔, 𝜔2}

a) Associative law holds 1 × 𝜔 × 𝜔2 = 1 × 𝜔 × 𝜔2 1 × 𝜔3 = 𝜔 × 𝜔2 𝜔3 = 𝜔3

1 = 1 ∈ 𝐺 b) 1 is the identity element with respect to multiplication

As, 1 × 1 = 1 ∈ 𝐺 1 × 𝜔 = 𝜔 ∈ 𝐺 1 × 𝜔2 = 𝜔2 ∈ 𝐺

c) Each element has an inverse in G Inverse of 1 is 1∈ 𝐺 Inverse of 𝜔 is 𝜔2 ∈ 𝐺 Inverse of 𝜔2 is 𝜔 ∈ 𝐺

𝐺, . is a group For Abelian:

1 × 𝜔 = 𝜔 × 1 = 𝜔 ∈ 𝐺 1 × 𝜔2 = 𝜔2 × 1 = 𝜔2 ∈ 𝐺

𝜔 × 𝜔2 = 𝜔2 × 𝜔 = 𝜔3 = 1 ∈ 𝐺 ‘.’ Is multiplication 𝐺, . is a Abelian group

Q.3(ii) QUESTION: If three books are picked at random from a shelf containing 3 novels, 4 book of poems and a dictionary. What is the probability that: (i) dictionary is selected (ii) one novel and 2 book of poems are selected SOLUTION: Total novels = 3 Total book of poems = 4 Total dictionary = 1 Total books = 3 + 4 + 1 = 8 Three books are selected at random 𝑂 𝑆 = 𝐶3.

8

(i) dictionary is selected Let A be the event that a dictionary is selected

𝑃 𝐴 =𝑂 𝐴

𝑂 𝑆

=𝐶2.

7 𝐶1.1

𝐶3.8

=21

56

=3

8

(ii) one novel and 2 book of poems are selected Let B be the event that one novel and 2 book of poems are selected

𝑃 𝐵 =𝑂 𝐵

𝑂 𝑆

=𝐶1.

3 𝐶2.4

𝐶3.8

=3 × 6

56

=9

28

OR Q.3(ii)

QUESTION: In how many ways can a party of 5 students and 2 teachers be formed out of 15 students and 5 teachers. SOLUTION: Total students = 15 Total teachers = 5 A party of 5 students and 2 teachers be formed 𝑇𝑜𝑡𝑎𝑙 𝑤𝑎𝑦𝑠 = 𝐶5.

15 𝐶2.5

= 3003 × 10 = 30030

Q.3(iii) QUESTION: Prove by the principle of Mathematical Induction. 1

1.2+

1

2.3+

1

3.4+. . . +

1

𝑛(𝑛+1)=

𝑛

𝑛+1, ∀𝑛 ∊ ℕ

SOLUTION: PROOF I:

Verifying p(n) for 𝑛 = 1 1

1.2=

1

1+1

1

2=

1

2 Verified

𝑃 𝑛 𝑖𝑠 𝑡𝑟𝑢𝑒 𝑓𝑜𝑟 𝑛 = 1

PROOF II:

Assuming that p(𝑛) is true for 𝑛 = 𝑘

So we have, 1

1.2+

1

2.3+

1

3.4+. . . +

1

𝑘(𝑘+1)=

𝑘

𝑘+1 ----(1)

Kth term =1

𝑘(𝑘 + 1)

k + 1 th term =1

(𝑘 + 1)(𝑘 + 2)

Adding 1

(𝑘+1)(𝑘+2) both sides in (1)

1

1.2+

1

2.3+

1

3.4+. . . +

1

𝑘 𝑘+1 +

1

𝑘+1 𝑘+2 =

𝑘

𝑘+1 +

1

𝑘+1 𝑘+2

=𝑘 𝑘+2 +1

𝑘+1 𝑘+2

=𝑘2+2𝑘+1

𝑘+1 𝑘+2

= 𝑘+1 2

𝑘+1 𝑘+2

=𝑘+1

𝑘+1+1

𝑊𝑕𝑖𝑐𝑕 𝑖𝑠 𝑡𝑟𝑢𝑒 𝑓𝑜𝑟 𝑛 = 𝑘 + 1

𝐻𝑒𝑛𝑐𝑒, 𝑃 𝑛 𝑖𝑠 𝑡𝑢𝑟𝑒 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑛𝑎𝑡𝑢𝑟𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟𝑠 𝑛.

OR Q.3(iii)

QUESTION: Find the sum of the following series: 212 + 222 + 232 + . . . + 502 SOLUTION: 212 + 222 + 232 + . . . + 502

= 502 − 212

∵ 𝑛2 =𝑛 𝑛 + 1 (2𝑛 + 1)

6

=50 50 + 1 (2 × 50 + 1)

6−

21 21 + 1 (2 × 21 + 1)

6

=50 51 (101)

6−

21 22 (41)

6

= 42925 − 3157 = 39768

Q.3(iv) QUESTION: Find the sum of an A.P., of nineteen terms whose middle term in 10. SOLUTION: Number of terms =n=19

𝑀𝑖𝑑𝑑𝑙𝑒 𝑡𝑒𝑟𝑚 =𝑛+1

2𝑡𝑕 𝑡𝑒𝑟𝑚

𝑀. 𝑇 =20

2= 10𝑡𝑕 𝑡𝑒𝑟𝑚

In an A.P. nth term = Tn=a+(n-1)d 𝑇10 = 𝑎 + 10 − 1 𝑑 10 = 𝑎 + 9𝑑 𝑎 + 9𝑑 = 10 ---(1) In an A.P.

Sum of n terms = 𝑆𝑛 =𝑛

2{2𝑎 + 𝑛 − 1 𝑑}

Sum of 19 terms = 𝑆19 =19

2{2𝑎 + 19 − 1 𝑑}

𝑆19 =19

2{2𝑎 + 18𝑑}

𝑆19 =19

2× 2 𝑎 + 9𝑑

𝑆19 = 19 10

𝑆19 = 190 [𝑈𝑠𝑖𝑛𝑔 (1)]

Q.3(v) QUESTION:

Find the value of n so that 𝑎𝑛+1+𝑏𝑛+1

𝑎𝑛 +𝑏𝑛 may become the H.M. between a and b.

SOLUTION:

H.M. between a and b is 2𝑎𝑏

𝑎+𝑏

According to the question: 𝑎𝑛+1 + 𝑏𝑛+1

𝑎𝑛 + 𝑏𝑛=

2𝑎𝑏

𝑎 + 𝑏

𝑎 + 𝑏 𝑎𝑛+1 + 𝑏𝑛+1 = 2𝑎𝑏 𝑎𝑛 + 𝑏𝑛 𝑎𝑛+2 + 𝑎𝑏𝑛+1 + 𝑎𝑛+1𝑏 + 𝑏𝑛+2 = 2𝑎𝑛+1𝑏 + 2𝑎𝑏𝑛+1 𝑎𝑛+2 + 𝑎𝑏𝑛+1 − 2𝑎𝑏𝑛+1 + 𝑎𝑛+1𝑏 − 2𝑎𝑛+1𝑏 + 𝑏𝑛+2 = 0 𝑎𝑛+2 − 𝑎𝑏𝑛+1 − 𝑎𝑛+1𝑏 + 𝑏𝑛+2 = 0 𝑎𝑎𝑛+1 − 𝑎𝑛+1𝑏 − 𝑎𝑐 + 𝑏𝑏𝑛+1 = 0 𝑎𝑛+1 𝑎 − 𝑏 —𝑏𝑛+1 𝑎 − 𝑏 = 0 𝑎 − 𝑏 𝑎𝑛+1 − 𝑏𝑛+1 = 0 𝑎𝑛+1 − 𝑏𝑛+1 = 0 𝑎𝑛+1 = 𝑏𝑛+1 𝑎𝑛+1

𝑏𝑛+1= 1

𝑎

𝑏 𝑛+1

= 𝑎

𝑏

0

By comparing 𝑛 + 1 = 0

𝑛 = −1

OR Q.3(v)

QUESTION: Find the first term of a G.P., whose second term is 3 and sum to infinity is 12. SOLUTION: In a G.P.: 𝑇𝑛 = 𝑎𝑟𝑛−1 𝑇2 = 𝑎𝑟2−1 = 3 𝑎𝑟 = 3 --- 1

𝑆 =𝑎

1 − 𝑟 = 12

𝑎

1 − 𝑟 = 12

× 𝑖𝑛𝑔 𝑏𝑦 𝑟, 𝑏𝑜𝑡𝑕 𝑠𝑖𝑑𝑒𝑠 𝑎𝑟

1 − 𝑟 = 12𝑟

3

1 − 𝑟 = 12𝑟 [𝑢𝑠𝑖𝑛𝑔 (1)]

1

1 − 𝑟 = 4𝑟

1 = 4𝑟 1 − 𝑟 1 = 4𝑟 − 4𝑟2 4𝑟2 − 4𝑟 + 1 = 0 2𝑟 − 1 2 = 0 2𝑟 − 1 = 0

𝑟 =1

2

1 ⟹ 𝑎𝑟 = 3

𝑎 1

2 = 3

𝑎 = 6

SECTION B

TRIGONOMETRY

Q.4(i) QUESTION: A belt 24.75 meters long passes around a 3.5 cm diameter pulley. As the belt makes 3 complete revolutions in a minute, how many radians does the wheel turn in two second? SOLUTION: Length of belt = 24.75𝑚

Diameter of the wheel of Pulley = 3.5𝑐𝑚

∵ 𝑟 =𝐷𝑖𝑎𝑚𝑒𝑡𝑒𝑟

2

𝑟 =3.5

2𝑐𝑚 = 1.75𝑐𝑚

𝑟 =1.75

100= 0.0175𝑚

As the belt makes three complete revolutions in a minute

𝑠 = 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑐𝑜𝑣𝑒𝑟𝑒𝑑 𝑏𝑦 𝑡𝑕𝑒 𝑏𝑒𝑙𝑡 𝑖𝑛 𝑎 𝑚𝑖𝑛𝑢𝑡𝑒 = 3(24.75)

= 74.25𝑚

𝑠 = 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑐𝑜𝑣𝑒𝑟𝑒𝑑 𝑏𝑦 𝑡𝑕𝑒 𝑏𝑒𝑙𝑡 𝑖𝑛 𝑎 𝑠𝑒𝑐𝑜𝑛𝑑 =74.25

60

𝑠 = 1.2375𝑚

∵ 𝑠 = 𝑟𝜃

𝜃 =𝑠

𝑟=

1.2375

0.0175= 70.714

For 2 seconds 𝜃 = 70.714 × 2

𝜃 = 141.428𝑟𝑎𝑑𝑖𝑎𝑛

𝑠𝑒𝑐𝑜𝑛𝑑

Q.4(ii) QUESTION: Find the period of 𝑡𝑎𝑛𝑥. SOLUTION: 𝐿𝑒𝑡 𝑓 𝑥 = 𝑡𝑎𝑛𝑥 𝑓 𝑥 + 𝑝 = 𝑡𝑎𝑛(𝑥 + 𝑝)

=𝑡𝑎𝑛𝑥 + 𝑡𝑎𝑛𝑝

1 − 𝑡𝑎𝑛𝑥𝑡𝑎𝑛𝑝

Replacing p with 𝜋, we get

𝑓 𝑥 + 𝜋 =𝑡𝑎𝑛𝑥 + 𝑡𝑎𝑛𝜋

1 − 𝑡𝑎𝑛𝑥𝑡𝑎𝑛𝜋

= 𝑡𝑎𝑛𝑥 + 0

1 − 𝑡𝑎𝑛𝑥(0)

= 𝑡𝑎𝑛𝑥

1 − 0

= 𝑡𝑎𝑛𝑥 𝑓 𝑥 + 𝜋 = 𝑓(𝑥)

𝐻𝑒𝑛𝑐𝑒, 𝑓 𝑥 𝑖𝑠 𝑡𝑕𝑒 𝑝𝑒𝑟𝑖𝑜𝑑𝑖𝑐 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝑝𝑒𝑟𝑖𝑜𝑑 𝜋.

Q.4(iii) QUESTION: If a=b=c then prove that r: R : r1 = 1 : 2 : 3, where r, R, r1 have their usual meanings. SOLUTION: ? ? r: R ∶ r1 = 1 ∶ 2 ∶ 3

𝑟 =∆

𝑠, 𝑅 =

𝑎𝑏𝑐

4∆, 𝑟1 =

∆

𝑠 − 𝑎

Given a=b=c

∆= 𝑆 𝑆 − 𝑎 𝑆 − 𝑏 𝑆 − 𝑐 ----(1)

Where, 𝑆 =𝑎+𝑏+𝑐

2

Here, 𝑆 =𝑎+𝑎+𝑎

2=

3𝑎

2

𝑆 − 𝑎 = 𝑆 − 𝑏 = 𝑆 − 𝑐 =3𝑎

2−

𝑎

1=

𝑎

2

𝑆 − 𝑎 =𝑎

2, 𝑆 − 𝑏 =

𝑎

2, 𝑆 − 𝑐 =

𝑎

2

1 ⇒ ∆= 3𝑎

2×

𝑎

2×

𝑎

2×

𝑎

2

∆= 3𝑎4

16

∆= 3 𝑎2

4

Now, 𝑟 =∆

𝑠=

3 𝑎2

4

3𝑎

2

= 3 𝑎2

4×

2

3𝑎=

3𝑎

2×3 =

3𝑎

6

𝑅 =𝑎𝑏𝑐

4∆=

𝑎. 𝑎. 𝑎

4 × 3 𝑎2

4

=𝑎

3=

𝑎

3×

3

3=

3𝑎

3

𝑟1 =∆

𝑠 − 𝑎=

3 𝑎2

4

𝑎2

= 3 𝑎2

4×

2

𝑎=

3𝑎

2

Now, 𝑟: 𝑅: 𝑟1 = 3𝑎

6∶

3𝑎

3 ∶

3𝑎

2

𝑋𝑖𝑛𝑔 𝑅. 𝐻. 𝑆 𝑏𝑦 6

3𝑎, 𝑤𝑒 𝑔𝑒𝑡

𝑟: 𝑅: 𝑟1 = 3𝑎

6 ×

6

3𝑎:

3𝑎

3×

6

3𝑎 ∶

3𝑎

2×

6

3𝑎

𝑟: 𝑅: 𝑟1 = 1 ∶ 2 ∶ 3

Q5(iv) QUESTION: 𝑡𝑎𝑛2𝜃𝑐𝑜𝑡𝜃 = 3 SOLUTION: 𝑡𝑎𝑛2𝜃𝑐𝑜𝑡𝜃 = 3

2𝑡𝑎𝑛𝜃

1 − tan2 𝜃×

1

𝑡𝑎𝑛𝜃= 3

2

1 − tan2 𝜃= 3

2 = 3 1 − tan2 𝜃 2 = 3 − 3 tan2 𝜃 3 tan2 𝜃 = 3 − 2 3 tan2 𝜃 = 1

tan2 𝜃 =1

3

𝑡𝑎𝑛𝜃 = ±1

3

Either:

𝑡𝑎𝑛𝜃 =1

3

Or:

𝑡𝑎𝑛𝜃 = −1

3

𝜃 = tan−1 1

3

𝜃 =𝜋

6

𝜃 = tan−1 −1

3

𝜃 = −𝜋

6

𝐺. 𝑆 = 2𝑛𝜋 +𝜋

6 ∪ 2𝑛𝜋 −

𝜋

6

𝑛 ∈ ℤ

Q.4(v) QUESTION: Prove that: 𝑇𝑎𝑛−1 1

5+ 𝑇𝑎𝑛−1 1

4= 𝑇𝑎𝑛−1 9

19

SOLUTION:

? ? 𝑇𝑎𝑛−11

5+ 𝑇𝑎𝑛−1

1

4= 𝑇𝑎𝑛−1

9

19

𝐿𝑒𝑡 𝑦 = 𝑇𝑎𝑛−1 1

5+ 𝑇𝑎𝑛−1 1

4= 𝐴 + 𝐵 ------(1)

Where, 𝐴 = 𝑇𝑎𝑛−1 1

5 and 𝐵 = 𝑇𝑎𝑛−1 1

4

So, 𝑇𝑎𝑛𝐴 =1

5 and 𝑇𝑎𝑛𝐵 =

1

4

1 ⇒ 𝑦 = 𝐴 + 𝐵 Taking Tan of both sides 𝑇𝑎𝑛𝑦 = 𝑇𝑎𝑛(𝐴 + 𝐵)

𝑇𝑎𝑛𝑦 =𝑇𝑎𝑛𝐴 + 𝑇𝑎𝑛𝐵

1 − 𝑇𝑎𝑛𝐴 𝑇𝑎𝑛𝐵

𝑇𝑎𝑛𝑦 =

15

+14

1 −15

×14

=

4 + 520

1 −1

20 =

9201920

𝑇𝑎𝑛𝑦 =9

19

𝑦 = 𝑇𝑎𝑛−19

19

𝑇𝑎𝑛−1 1

5+ 𝑇𝑎𝑛−1 1

4= 𝑇𝑎𝑛−1 9

19 𝐹𝑟𝑜𝑚 1

OR Q.4(v)

QUESTION:

Prove that: 𝑆𝑖𝑛−1𝐴 + 𝑆𝑖𝑛−1𝐵 = sin−1 𝐴 1 − 𝐵2 + 𝐵 1 − 𝐴2

SOLUTION: 𝐿𝑒𝑡 𝑡 = 𝑆𝑖𝑛−1𝐴 + 𝑆𝑖𝑛−1𝐵 = 𝑥 + 𝑦 ------(1) Where, 𝑥 = 𝑆𝑖𝑛−1𝐴 and 𝑦 = 𝑆𝑖𝑛−1𝐵

So, 𝑆𝑖𝑛𝑥 = 𝐴 𝑆𝑖𝑛2𝑥 + cos2 𝑥 = 1 𝐴2 + cos2 𝑥 = 1 cos2 𝑥 = 1 − 𝐴2

𝐶𝑜𝑠𝑥 = 1 − 𝐴2

and 𝑠𝑖𝑛𝑦 = 𝐵

𝑆𝑖𝑛2𝑦 + cos2 𝑦 = 1 𝐵2 + cos2 𝑦 = 1 cos2 𝑦 = 1 − 𝐵2

𝐶𝑜𝑠𝑦 = 1 − 𝐵2

1 ⇒ 𝑡 = 𝑥 + 𝑦 Taking Sin of both sides 𝑠𝑖𝑛𝑡 = sin(𝑥 + 𝑦) 𝑠𝑖𝑛𝑡 = sinxcosy + cosxsiny

𝑠𝑖𝑛𝑡 = A 1 − 𝐵2 + B 1 − 𝐴2

𝑡 = sin−1 𝐴 1 − 𝐵2 + 𝐵 1 − 𝐴2 Proved.

SECTION C Q.5(a) QUESTION:

SOLUTION: Let five shares in A.P. are: 𝑎 − 4𝑑, 𝑎 − 2𝑑, 𝑎, 𝑎 + 2𝑑, 𝑎 + 4𝑑 A/c to the Ist condition: 𝑎 − 4𝑑 + 𝑎 − 2𝑑 + 𝑎 + 𝑎 + 2𝑑 + 𝑎 + 4𝑑 = 600 5𝑎 = 600

𝑎 = 120 A/c to the 2nd condition:

𝑎 − 4𝑑 + 𝑎 − 2𝑑 =1

7 𝑎 + 𝑎 + 2𝑑 + 𝑎 + 4𝑑

2𝑎 − 6𝑑 =1

7 3𝑎 + 6𝑑

14𝑎 − 42𝑑 = 3𝑎 + 6𝑑 14𝑎 − 3𝑎 = 42𝑑 + 6𝑑 11𝑎 = 48𝑑 ⇒ 48𝑑 = 11𝑎

𝑑 =11 × 120

48=

55

2

Five shares are when a=120 and d=55

2:

𝑎 − 4𝑑, 𝑎 − 2𝑑, 𝑎, 𝑎 + 2𝑑, 𝑎 + 4𝑑

120 − 4 55

2 , 120 − 2

55

2 , 120, 120 + 2

55

2 , 120 + 4

55

2

10, 65, 120, 175, 230

OR Q.5(a)

QUESTION: Note: BIEK has made mistake here, by 35th term, it is impossible question. It is 34th term.

𝑾𝒓𝒐𝒏𝒈 𝑸𝒖𝒆𝒔𝒕𝒊𝒐𝒏: 𝐼𝑛 𝑎𝑛 𝐻. 𝑃. , 𝑡𝑕𝑒 10𝑡𝑕 𝑡𝑒𝑟𝑚 𝑖𝑠 35, 35𝑡𝑕 𝑡𝑒𝑟𝑚 𝑖𝑠 25.𝐼𝑓 𝑡𝑕𝑒 𝑙𝑎𝑠𝑡 𝑡𝑒𝑟𝑚 𝑖𝑠 2, 𝑓𝑖𝑛𝑑 𝑡𝑕𝑒 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑡𝑒𝑟𝑚𝑠.

SOLUTION: In an H.P:

pth term = 10th term = 𝑥 = 35, 𝑝 = 10

qth term = 35th term = 𝑦 = 25, 𝑞 = 35

rth term = last term = 𝑧 = 2, 𝑟 == 𝑛 =?

We have,

1

𝑥

1

𝑦

1

𝑧

𝑝 𝑞 𝑟1 1 1

= 0

1

35

1

25

1

210 35 𝑛1 1 1

= 0

×ing 𝑅1 by 350, we get

10 14 17510 35 𝑛1 1 1

= 0 × 350

Expanding by 𝐶3, we get

175 10 351 1

− 𝑛 10 141 1

+ 1 10 1410 35

= 0

175 10 − 35 − 𝑛 10 − 14 + 250 − 140 = 0 175 −25 − 𝑛 −4 + 350 − 140 = 0 −4375 + 4𝑛 + 210 = 0 4𝑛 − 4165 = 0 4𝑛 = 4165

𝑧 =4165

4 Not a positive integer

Hence, wrong Question

Q.5(b) QUESTION:

Prove law of Cosine: a2 =b

2 + c

2 – 2bccos𝛼

PROOF

Law of Cosine: c2 =a

2 + b

2 – 2abcos𝜸

Proof: We place a ∆ABC in x-, y- coordinate system such that C(0,0) is at the origin and B(a,0) on

positive x-axis as shown in the figure.

As, cos 1800 − 𝛾 =𝑏𝑎𝑠𝑒

𝑝𝑒𝑟=

𝐶𝐿

𝐴𝐶

cos 1800 𝑐𝑜𝑠𝛾 + sin 1800 𝑠𝑖𝑛𝛾 =𝐶𝐿

𝑏

−1 𝑐𝑜𝑠𝛾 + (0)𝑠𝑖𝑛𝛾 =𝐶𝐿

𝑏

𝑐𝑜𝑠𝛾 =𝐶𝐿

𝑏

𝐶𝐿 = −𝑏𝑐𝑜𝑠𝛾

As, sin 1800 − 𝛾 =𝑝𝑒𝑟

𝑝𝑒𝑟=

𝐴𝐿

𝐴𝐶

sin 1800 𝑐𝑜𝑠𝛾 − cos 1800 𝑠𝑖𝑛𝛾 =𝐶𝐿

𝑏

0 𝑐𝑜𝑠𝛾 − (−1)𝑠𝑖𝑛𝛾 =𝐶𝐿

𝑏

𝑠𝑖𝑛𝛾 =𝐶𝐿

𝑏

𝐶𝐿 = 𝑏𝑠𝑖𝑛𝛾

So, coordinates of A are (b.cos𝛾, b.sin𝛾)

We have distance formula as:

d = (𝑥2 − 𝑥1)2 + (𝑦2 − 𝑦1)2

Here 𝐴𝐵 = c = (𝑏𝑐𝑜𝑠𝛾 − 𝑎)2 + (𝑏𝑠𝑖𝑛𝛾 − 0)2

c = 𝑏2𝑐𝑜𝑠2𝛾 − 2𝑎𝑏𝑐𝑜𝑠𝛾 + 𝑎2 + 𝑏2𝑠𝑖𝑛2𝛾

c = 𝑏2𝑐𝑜𝑠2𝛾 + 𝑏2𝑠𝑖𝑛2𝛾 − 2𝑎𝑏𝑐𝑜𝑠𝛾 + 𝑎2

c = 𝑏2(𝑐𝑜𝑠2𝛾 + 𝑠𝑖𝑛2𝛾) − 2𝑎𝑏𝑐𝑜𝑠𝛾 + 𝑎2

c = 𝑏2(1) − 2𝑎𝑏𝑐𝑜𝑠𝛾 + 𝑎2 Squaring both sides

c2 = 𝑏2 − 2𝑎𝑏𝑐𝑜𝑠𝛾 + 𝑎2

And hence, c2 = a

2 + b

2 – 2abcos𝛾

Similarly: a

2 = b

2 + c

2 – 2bccos𝛼

b2 = a

2 + c

2 – 2accos𝛽

OR Q.5(b)

QUESTION: Prove fundamental law: cos(𝛼 − 𝛽) = cos𝛼. cos𝛽 + sin𝛼. sin𝛽 PROOF:

Fundamental Law: Consider a unit circle with centre at O(0,0) as shown in figure.

Let P(cos𝛽, sin𝛽) and Q(cos𝛼, sin𝛼) be any two points in unit circle.

We have distance formula as:

d = (𝑥2 − 𝑥1)2 + (𝑦2 − 𝑦1)2

Here 𝑃𝑄 = (𝑐𝑜𝑠𝛼 − 𝑐𝑜𝑠𝛽)2 + (𝑠𝑖𝑛𝛼 − 𝑠𝑖𝑛𝛽)2 ------------------(1)

Now rotate the axes so that the positive direction of X-axis passes through the point P. Then with respect to this coordinate system, the coordinates of P and Q become (1,0) and (cos(𝛼 − 𝛽), sin(𝛼 − 𝛽)) respectively.

So, 𝑃𝑄 = [ cos 𝛼 − 𝛽 − 1]2 + [ sin 𝛼 − 𝛽 − 0 ]2 ------------------(2)

Comparing (1) and (2), we get

(𝑐𝑜𝑠𝛼 − 𝑐𝑜𝑠𝛽)2 + (𝑠𝑖𝑛𝛼 − 𝑠𝑖𝑛𝛽)2 = [ cos 𝛼 − 𝛽 − 1]2 + [ sin 𝛼 − 𝛽 − 0 ]2

or (𝑐𝑜𝑠𝛼 − 𝑐𝑜𝑠𝛽)2 + (𝑠𝑖𝑛𝛼 − 𝑠𝑖𝑛𝛽)2 = [ cos 𝛼 − 𝛽 − 1]2 + [ sin 𝛼 − 𝛽 − 0 ]2

or cos2𝛼 – 2cos𝛼.cos𝛽 + cos2𝛽 + sin2𝛼 – 2sin𝛼.sin𝛽 + sin2𝛽 = cos2(𝛼 − 𝛽) – 2cos(𝛼 − 𝛽) + 1 + sin2(𝛼 − 𝛽)

or sin2𝛼 + cos2𝛼 – 2cos𝛼.cos𝛽 – 2sin𝛼.sin𝛽 + sin2𝛽 + cos2𝛽 = sin2(𝛼 − 𝛽)+ cos2(𝛼 − 𝛽) + 1 – 2cos(𝛼 − 𝛽)

or 1 – 2cos𝛼.cos𝛽 – 2sin𝛼.sin𝛽 + 1 = 1 + 1 – 2cos(𝛼 − 𝛽)

or – 2cos𝛼.cos𝛽 – 2sin𝛼.sin𝛽 = – 2cos(𝛼 − 𝛽) Dividing by -2, we get or cos𝛼.cos𝛽 + sin𝛼.sin𝛽 = cos(𝛼 − 𝛽)

Hence, cos(𝛼 − 𝛽) = cos𝛼.cos𝛽 + sin𝛼.sin𝛽

Q.6(a) QUESTION:

Show that: 2 = 1 +1

22 +1.3

2!.24 +1.3.5

3!.26 + . . .

SOLUTION:

2 = 1 +1

22 +1.3

2!.24 +1.3.5

3!.26 + . . .

Comparing R.H.S with R.H.S of

1 + 𝑥 𝑛 = 1 + 𝑛𝑥 +𝑛 𝑛 − 1

2!𝑥2 + . . .

𝑛𝑥 =1

22=

1

4 ‐ ‐ ‐ (1)

Squaring both sides

𝑛2𝑥2 =1

16 ‐ ‐ ‐ (2)

𝑛 𝑛 − 1

2!𝑥2 =

1.3

2!. 24

𝑛 𝑛 − 1 𝑥2 =3.2!

2! .16

𝑛 𝑛 − 1 𝑥2 =3

16

÷ 𝑖𝑛𝑔 𝑏𝑦 2 , 𝑤𝑒 𝑔𝑒𝑡 𝑛 𝑛 − 1 𝑥2

𝑛2𝑥2=

3

16÷

1

16

𝑛 − 1

𝑛=

3

16× 16

𝑛 − 1

𝑛= 3 ⟹ 𝑛 − 1 = 3𝑛

−1 = 3𝑛 − 𝑛 2𝑛 = −1

𝑛 = −1

2

1 ⟹ 𝑛𝑥 =1

4

−1

2𝑥 =

1

4

𝑥 = −1

2

Now,

𝑅. 𝐻. 𝑆 = 1 −1

2

−12

= 1

2 −

12

= 2 12

= 2 = 𝐿. 𝐻. 𝑆

Q.6(b) QUESTION: Solve the following system of equations by Cramer’s Rule:

𝑥 + 𝑦 + 𝑧 = 𝑑, 𝑥 + 1 + 𝑑 𝑦 + 𝑧 = 2𝑑, 𝑥 + 𝑦 + (1 + 𝑑)𝑧 = 0 SOLUTION:

𝑥 + 𝑦 + 𝑧 = 𝑑 𝑥 + 1 + 𝑑 𝑦 + 𝑧 = 2𝑑 𝑥 + 𝑦 + 1 + 𝑑 𝑧 = 0 𝑑 ≠ 0 The determinant of the coefficient is

𝐴 = 1 1 11 1 + 𝑑 11 1 1 + 𝑑

Expanding by 𝑅1

= 1 1 + 𝑑 1

1 1 + 𝑑 − 1

1 11 1 + 𝑑

+ 1 1 1 + 𝑑1 1

= 1 + 𝑑 2 − 1 − 1 + 𝑑 − 1 + 1 − (1 + 𝑑) = 1 + 2𝑑 + 𝑑2 − 1 − 𝑑 − 𝑑 𝐴 = 𝑑2 ≠ 0, Non-singular

𝐴𝑥 = 𝑑 1 1

2𝑑 1 + 𝑑 10 1 1 + 𝑑

Expanding by 𝐶1

= 𝑑 1 + 𝑑 1

1 1 + 𝑑 − 2𝑑

1 11 1 + 𝑑

+ 0

= 𝑑 1 + 𝑑 2 − 1 − 2𝑑 1 + 𝑑 − 1 + 0 = 𝑑 1 + 2𝑑 + 𝑑2 − 1 − 2𝑑(𝑑) = 𝑑 2𝑑 + 𝑑2 − 2𝑑2 = 2𝑑2 + 𝑑3 − 2𝑑2

𝐴𝑥 = 𝑑3

𝐴𝑦 = 1 𝑑 11 2𝑑 11 0 1 + 𝑑

Expanding by 𝑅3

= 1 𝑑 1

2𝑑 1 − 0 + (1 + 𝑑)

1 𝑑1 2𝑑

= 𝑑 − 2𝑑 + 1 + 𝑑 2𝑑 − 𝑑 = −𝑑 + 1 + 𝑑 𝑑 = −𝑑 + 𝑑 + 𝑑2

𝐴𝑦 = 𝑑2

𝐴 = 1 1 𝑑1 1 + 𝑑 2𝑑1 1 0

Expanding by 𝑅3

= 1 1 𝑑

1 + 𝑑 2𝑑 − 1

1 𝑑1 2𝑑

+ 0

= 2𝑑 − 𝑑(1 + 𝑑) − 2𝑑 − 𝑑 = 2𝑑 − 𝑑 − 𝑑2 − 𝑑

𝐴𝑧 = −𝑑2

By Cramer’s rule

𝑥 =|𝐴𝑥 |

𝐴 =

𝑑3

𝑑2= 𝑑

𝑦 ==|𝐴𝑦 |

|𝐴|=

𝑑2

𝑑2= 1

𝑧 ==|𝐴𝑧 |

|𝐴|=

−𝑑2

𝑑2= −1

The solution is: 𝑥 = 𝑑, 𝑦 = 1, 𝑧 = −1

Q.7(a) QUESTION: Find the remaining trigonometric functions using radian function if 𝑠𝑖𝑛𝜃 = 0.6 and 𝑡𝑎𝑛𝜃 is negative. SOLUTION: As, 𝑠𝑖𝑛𝜃 is positive and 𝑡𝑎𝑛𝜃 is negative so 𝜌(𝜃) is in second quadrant

In a unit circle

𝑥2 + 𝑦2 = 1 ----(1)

Where 𝑥 = 𝑐𝑜𝑠𝜃 and 𝑦 = 𝑠𝑖𝑛𝜃

Given, sin𝜃 = 𝑦 = 0.6

𝑦 = 0.6 ---(2)

(1) ⇒ 𝑥2 + 0.6 2 = 1

𝑥2 + 0.36 = 1

𝑥2 = 1 − 0.36

𝑥2 = 0.64

𝑥 = ±0.8

As 𝜌(𝜃) is in second quadrant

𝑥 = −0.8

𝑠𝑖𝑛𝜃 = 𝑦 = 0.6

𝑐𝑜𝑠𝑒𝑐𝜃 =1

𝑦=

1

0.6=

𝑐𝑜𝑠𝜃 = 𝑥 = −0.8

𝑠𝑒𝑐𝜃 =1

𝑥=

−1

0.8=

𝑡𝑎𝑛𝜃 =𝑦

𝑥=

0.6

−0.8=

𝑐𝑜𝑡𝜃 =𝑥

𝑦=

Q.7(a) QUESTION: Prove any two of the following:

(i) 1−𝑠𝑖𝑛𝜃

1−𝑠𝑖𝑛𝜃= 𝑠𝑒𝑐𝜃 − 𝑡𝑎𝑛𝜃

(ii) 𝑠𝑖𝑛2𝜃

𝑠𝑖𝑛𝜃−

𝑐𝑜𝑠2𝜃

𝑐𝑜𝑠𝜃= 𝑠𝑒𝑐𝜃

(iii) 𝑠𝑖𝑛7𝜃−𝑠𝑖𝑛5𝜃

𝑐𝑜𝑠7𝜃+𝑐𝑜𝑠5𝜃= 𝑡𝑎𝑛𝜃 OR 𝑡𝑎𝑛

𝜃

2 =

𝑠𝑖𝑛𝜃

1+𝑐𝑜𝑠𝜃

SOLUTION:

(i) 1−𝑠𝑖𝑛𝜃

1+𝑠𝑖𝑛𝜃= 𝑠𝑒𝑐𝜃 − 𝑡𝑎𝑛𝜃

𝐿. 𝐻. 𝑆 = 1−𝑠𝑖𝑛𝜃

1+𝑠𝑖𝑛𝜃

× 𝑖𝑛𝑔 𝑎𝑛𝑑 ÷ 𝑖𝑛𝑔 𝑏𝑦 1 − 𝑠𝑖𝑛𝜃 , 𝑤𝑒 𝑔𝑒𝑡

= 1−𝑠𝑖𝑛𝜃

1+𝑠𝑖𝑛𝜃×

1−𝑠𝑖𝑛𝜃

1−𝑠𝑖𝑛𝜃

= 1−𝑠𝑖𝑛𝜃 2

1 2− 𝑠𝑖𝑛𝜃 2

= 1−𝑠𝑖𝑛𝜃 2

1−sin 2 𝜃

= 1−𝑠𝑖𝑛𝜃 2

cos 2 𝜃

=1−𝑠𝑖𝑛𝜃

𝑐𝑜𝑠𝜃=

1

𝑐𝑜𝑠𝜃−

𝑠𝑖𝑛𝜃

𝑐𝑜𝑠𝜃= 𝑠𝑒𝑐𝜃 − 𝑡𝑎𝑛𝜃 = 𝑅. 𝐻. 𝑆

(ii) 𝑠𝑖𝑛2

𝑠𝑖𝑛−

𝑐𝑜𝑠2

𝑐𝑜𝑠 = 𝑠𝑒𝑐

𝐿. 𝐻. 𝑆 = 𝑠𝑖𝑛2𝜃

𝑠𝑖𝑛𝜃−

𝑐𝑜𝑠2𝜃

𝑐𝑜𝑠𝜃

=𝑠𝑖𝑛2𝜃𝑐𝑜𝑠𝜃 −𝑐𝑜𝑠2𝜃𝑠𝑖𝑛𝜃

𝑠𝑖𝑛𝜃𝑐𝑜𝑠𝜃

=𝑠𝑖𝑛(2𝜃 − 𝜃)

𝑠𝑖𝑛𝜃𝑐𝑜𝑠𝜃

=𝑠𝑖𝑛𝜃

𝑠𝑖𝑛𝜃𝑐𝑜𝑠𝜃

=1

𝑐𝑜𝑠𝜃

= 𝑠𝑒𝑐𝜃 = 𝑅. 𝐻. 𝑆

(iii) 𝑠𝑖𝑛7θ –sin 5θ

𝑐𝑜𝑠7θ +cos 5θ= 𝑡𝑎𝑛θ

𝐿. 𝐻. 𝑆 =𝑠𝑖𝑛7𝜃 − 𝑠𝑖𝑛5𝜃

𝑐𝑜𝑠7𝜃 + 𝑐𝑜𝑠5𝜃

= 2 cos

7𝜃 + 5𝜃2 sin

7𝜃 − 5𝜃2

2 cos7𝜃 + 5𝜃

2cos

7𝜃 − 5𝜃2

∵ 𝑠𝑖𝑛𝑈 − 𝑠𝑖𝑛𝑉 = 2 cos

𝑈 + 𝑉

2𝑠𝑖𝑛

𝑈 − 𝑉

2

𝑐𝑜𝑠𝑈 + 𝑐𝑜𝑠𝑉 = 2 cos𝑈 + 𝑉

2𝑐𝑜𝑠

𝑈 − 𝑉

2

= cos

12𝜃2 sin

2𝜃2

cos12𝜃

2 cos2𝜃2

= cos 6𝜃𝑠𝑖𝑛𝜃

cos 6𝜃𝑐𝑜𝑠𝜃

=𝑠𝑖𝑛𝜃

𝑐𝑜𝑠𝜃

= 𝑡𝑎𝑛𝜃 = 𝑅. 𝐻. 𝑆 OR

(iii) 𝑡𝑎𝑛𝜃

2 =

𝑠𝑖𝑛𝜃

1+𝑐𝑜𝑠𝜃

𝑅. 𝐻. 𝑆 =sin 𝜃

1 + 𝑐𝑜𝑠𝜃

=2𝑠𝑖𝑛

𝜃

2𝑐𝑜𝑠

𝜃

2

1+ 𝑐𝑜𝑠 2𝜃

2 – 𝑐𝑜𝑠2𝜃

2

∵ sin 𝜃 = 2𝑠𝑖𝑛𝜃

2𝑐𝑜𝑠

𝜃

2 & 𝑐𝑜𝑠𝜃 = 𝑐𝑜𝑠2 𝜃

2 – 𝑐𝑜𝑠2 𝜃

2

=2𝑠𝑖𝑛

𝜃

2𝑐𝑜𝑠

𝜃

2

1−𝑐𝑜𝑠 2𝜃

2 + 𝑠𝑖𝑛2𝜃

2

=2𝑠𝑖𝑛

𝜃

2𝑐𝑜𝑠

𝜃

2

𝑠𝑖𝑛2𝜃

2 + 𝑠𝑖𝑛2𝜃

2

=2𝑠𝑖𝑛

𝜃

2𝑐𝑜𝑠

𝜃

2

2𝑠𝑖𝑛2𝜃

2

=𝑐𝑜𝑠

𝜃

2

𝑠𝑖𝑛𝜃

2

= 𝑡𝑎𝑛𝜃

2= 𝐿. 𝐻. 𝑆

Q.7(C) QUESTION: Solve: 𝑥2 + 𝑦2 = 34, 𝑥𝑦 + 15 = 0 SOLUTION:

𝑥2 + 𝑦2 = 34

𝑥𝑦 + 15 = 0

𝑥𝑦 = −15

𝑦 = −15

𝑥 ----(1)

2 ⟹ 𝑥2 + −15

𝑥

2

= 34 𝑈𝑠𝑖𝑛𝑔 (1)

𝑥2 +225

𝑥2 = 34

× 𝑖𝑛𝑔 𝑏𝑦 𝑥2 , 𝑤𝑒 𝑔𝑒𝑡

𝑥4 + 225 = 34𝑥2

𝑥4 − 34𝑥2 + 225 = 0

𝑥4 − 25𝑥2 − 9𝑥2 + 225 = 0

𝑥2 𝑥2 − 25 − 9 𝑥2 − 25 = 0

𝑥2 − 25 𝑥2 − 9 = 0

Either: 𝑥2 − 25 = 0

𝑥2 = 25 ⟹ 𝑥 = ±5

Either: 𝑥2 − 9 = 0

𝑥2 = 9 ⟹ 𝑥 = ±3

(1) ⟹ 𝑦 = −15

𝑥

𝑥 = 5

𝑦 = −15

5

𝑦 = −3, 𝑥 = 5

𝑥 = −5

𝑦 = −15

−5

𝑦 = 3, 𝑥 = −5

𝑥 = 3

𝑦 = −15

3

𝑦 = −5, 𝑥 = 3

𝑥 = −3

𝑦 = −15

−3

𝑦 = 5, 𝑥 = −3

𝑆. 𝑆 = 5, −3 , −5,3 , 3, −5 , (−3,5)