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,

cosec^{2}A – cot^{2}A = 1

⇒ cosec^{2} A = 1 + cot^{2} A

⇒ (cosec A)^{2} = cot^{2} 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,

sec^{2} A – tan^{2} A = 1

⇒ sec^{2} A = 1 + tan^{2} A

Question 2.

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

Solution—

sin^{2} A + cos^{2} A = 1

⇒ sin^{2} A = 1 – cos^{2} A

1 + tan^{2} A = sec^{2} A

tan^{2} A = sec^{2} A – 1

(tan A)^{2} = sec^{2} 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°

= sin^{2} 25° + cos^{2} 25°

= 1

Question 4.

Choose the correct option. Justify your choice.

(i) 9 sec^{2} A – 9 tan^{2} 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) sec^{2} A

(B) – 1

(C) cot^{2} A

(D) tan^{2} A

Solution:

(i) 9 sec^{2} A – 9 tan^{2} A

= 9 (sec^{2} A – tan^{2} 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 cosec^{2} A = 1 + cot^{2} 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 + tan^{2} A + cot^{2} A

(ix) (cosec A – sin A) (sec A – cos A) = \(\frac{1}{\tan A+\cot A}\)

[Hint : Simplify LHS and RHS separately]

(x) \(\left(\frac{1+\tan ^{2} A}{1+\cot ^{2} A}\right)\) = \(\left(\frac{1-\tan A}{1-\cot A}\right)^{2}\) = tan^{2} 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.

= {sin^{2} A + cosec^{2} A + 2} + {cos^{2} A + sec^{2} A + 2}

= 2 + 2 + (sin^{2}A + cos^{2} A) + sec^{2} A + cosec^{2} A

= 2 + 2 + 1 + 1 + tan^{2} A + 1 + cot^{2} A

[∵ sec^{2} A = tan ^{2} A + 1,

cosec^{2} A = cot^{2} A + 1]

= 7 + tan^{2}A + cot^{2} 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

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