RBSE Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.4

Rajasthan Board RBSE Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.4 Textbook Exercise Questions and Answers.

RBSE Class 10 Maths Solutions Chapter 8 Introduction to Trigonometry Exercise 8.4

Question 1.
Express the trigonometric ratios sin A, sec A and tan A in terms of cot A.
Solution—
Using the identity,
cosec²A – cot²A = 1
⇒ cosec² A = 1 + cot² A
⇒ (cosec A)² = cot² A + 1

Discarding negative values of sin A for acute angle A
So, sin A = \(\frac{1}{\sqrt{\cot ^{2} A+1}}\)
Using the Identity,
sec² A – tan² A = 1
⇒ sec² A = 1 + tan² A

Question 2.
Write all the other trigonometric ratios of ∠A in terms of sec A.
Solution—
sin² A + cos² A = 1
⇒ sin² A = 1 – cos² A

1 + tan² A = sec² A
tan² A = sec² A – 1
(tan A)² = sec² A – 1
∴ tan A = ±\(\sqrt{\sec ^{2} \mathrm{~A}-1}\) [Leaving – ve sign for acute angle 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:

(ii) sin 25° cos 65° + cos 25° sin 65°
= sin 25° × cos (90° – 25°) + cos 25° × sin (90° – 25°)
[∵ cos(90°- θ) = sinθ
sin(90°- θ) = cosθ
= sin 25° × sin 25° + cos 25° × cos 25°
= sin² 25° + cos² 25°
= 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) -1
(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) tan² A
Solution:
(i) 9 sec² A – 9 tan² A
= 9 (sec² A – tan² A) = 9 × 1 = 9
∴ Correct Option = (B)

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


∴ Correct Option = (D)

Question 5.
Prove the following identities, where the angles involved are acute angles for which the expressions are defined :
(i) (cosec θ – cot θ)² = \(\frac{1-\cos \theta}{1+\cos \theta}\)
(ii) \(\frac{\cos A}{1+\sin A}\) + \(\frac{1+\sin A}{\cos A}\) = 2 sec A
(iii) \(\frac{\tan \theta}{1-\cot \theta}\) + \(\frac{\cot \theta}{1-\tan \theta}\) = 1 + sec θ
[Hint : Write the expression in terms of sin θ and cos θ]
(iv) \(\frac{1+\sec A}{\sec A}\) = \(\frac{\sin ^{2} A}{1-\cos A}\)
[Hint : Simplify LHS and RHS separately]
(v) \(\frac{\cos A-\sin A+1}{\cos A+\sin A-1}\) = cosec A + cot A
using the identity cosec² A = 1 + cot² A.
(vi) \(\sqrt{\frac{1+\sin A}{1-\sin A}}\) = sec A + tan A
(vii) \(\frac{\sin \theta-2 \sin ^{3} \theta}{2 \cos ^{3} \theta-\cos \theta}\) = tan θ
(viii) (sin A + cosec A)² + (cos A + sec A)² = 7 + tan² A + cot² A
(ix) (cosec A – sin A) (sec A – cos A) = \(\frac{1}{\tan A+\cot A}\)
[Hint : Simplify LHS and RHS separately]
(x) \(\left(\frac{1+\tan ^{2} A}{1+\cot ^{2} A}\right)\) = \(\left(\frac{1-\tan A}{1-\cot A}\right)^{2}\) = tan² A
Solution:
(i) L.H.S = (cosec θ – cot θ)²






To remove the square root sign multiplying the numerator and the denominator by 1 + sin A.



= {sin² A + cosec² A + 2} + {cos² A + sec² A + 2}
= 2 + 2 + (sin²A + cos² A) + sec² A + cosec² A
= 2 + 2 + 1 + 1 + tan² A + 1 + cot² A
[∵ sec² A = tan ² A + 1,
cosec² A = cot² A + 1]
= 7 + tan²A + cot² A = R.H.S
∴ L.H.S = R.H.S Hence Proved

(ix) L.H.S = (cosec A – sin A) (sec A – cos A)




∴ L.H.S = R.H.S Hence Proved