Bihar Board Class 10th Maths Solutions Chapter 8 Introduction to Trigonometry Ex 8.4 Textbook Questions and Answers.

## BSEB Bihar Board Class 10th Maths Solutions Chapter 8 Introduction to Trigonometry Ex 8.4

Question 1.

Express the trigonometric ratios sin A, sec A and tan A in terms of cot A.

Solution:

Consider a ∆ ABC, in which ∠B = 90°.

Base = AB

Perp. = BC

and Hyp. = AC

∴ cot A = \(\frac{\text { Base }}{\text { Perp }}\) = \(\frac { AB }{ BC }\)

or \(\frac { AB }{ BC }\) = cot A = \(\frac { cot A }{ 1 }\)

Let AB = k cot A and BC = k.

Question 2.

Write the other trigonometric ratios of A in terms of sec A.

Solution:

Consider a ∆ ABC, in which ∠B = 90°

For ∠A, we have :

Base = AB,

Perp = BC

Hyp = AC.

∴ sec A = \(\frac{\text { Hyp }}{\text { Base }}\) = \(\frac { AC }{ AB }\)

or \(\frac { AC }{ AB }\) = sec A = \(\frac { sec A }{ 1 }\)

Let AB = k and AC = k sec A.

Question 3.

Evaluate :

(i) \(\frac{\sin ^{2} 63^{\circ}+\sin ^{2} 27^{\circ}}{\cos ^{2} 17^{\circ}+\cos ^{2} 73^{\circ}}\)

(ii) sin 25° cos 65° + cos 25° sin 65

Solution:

(i) Here, sin 63° = sin (90° – 27°) = cos 27°

and cos 17° = cos (90° – 73°) = sin 73°

(ii) sin 25° cos 65° + cos 25° sin 65°

= sin (90° – 65°) cos 65° + cos (90° – 65°) sin 65°

= cos 65° cos 65° + sin 65° sin 65°

= cos² 65° + sin² 65° = 1

Question 4.

Choose the correct option. Justify your choice :

(i) 9 sec² A – 9 tan² A =

(A) 1

(B) 9

(C) 8

(D) 0

(ii) (1 + tan θ + sec θ) (1 + cot θ – cosec θ) =

(A) 0

(B) 1

(C) 2

(D) none of these

(iii) (sec A + tan A)(1 – sin A) =

(A) sec A

(B) sin A

(C) cosec A

(D) cos A

(iv) \(\frac{1+\tan ^{2} A}{1+\cot ^{2} A}\)

(A) sec² A

(B) – 1

(C) cot² A

(D) none of these

Solution:

(i) (B), because

9 sec² A – 9 tan² A = 9 (sec² A – tan² A) = 9 x 1 = 9

(ii) (C), because

(1 + tan θ + sec θ )(1 + cot θ – cosec θ)

(iii) (D), because

(sec A + tan A)(1 – sin A) =

(iv) (D), because

Question 5.

Prove the following identities, where the angles involved are acute angles for which the expressions are defined :

Solution:

(i) We have :

L.H.S. = (cosec θ – cot θ)²

(ii) We have :

(ii) We have :

(iii) We have :

(iv) We have :

(v) We have :

(vi) We have :

(vii) We have :

(viii) We have :

L.H.S. = (sin A + cosec A)² + (cos A + sec A)²

= (sin² A + cosec² A + 2 sin A cosec A) + (cos² A + sec2 A + 2 cos A sec A)

= (sin² A + cosec² A + 2 sin A.\(\frac{1}{\sin A}\)) + (cos² A + sec2 A + 2 cos A. \(\frac{1}{\cos A}\))

= (sin²A + cosec² A + 2) + (cos² A + sec² A + 2)

= sin²A + cos² A + cosec² A + sec² A + 4

= 1 + (1 + cot² θ) + (1 + tan² A) + 7 + tan² A + cot² A [ ∵ cosec² A = 1 + cot² A and sec² A = 1 + tan² A]

= R.H.S.

(ix) We have :

(x) We have :