In this post, we will share NCERT Class 10th Maths Book Solutions Chapter 8 Introduction to Trigonometry Ex 8.4. These solutions are based on new NCERT Syllabus.

## NCERT 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:

Question 2.

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

Solution:

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

∵ sin 63° = sin (90° – 27°) = cos 27°

⇒ sin^{2} 63° = cos^{2} 27°

cos^{2} 73° = cos^{2} (90° – 17°) = sin^{2} 17°

∴ \(\frac{\sin ^{2} 63^{\circ}+\sin ^{2} 27^{\circ}}{\cos ^{2} 17^{\circ}+\cos ^{2} 73^{\circ}}=\frac{\cos ^{2} 27^{\circ}+\sin ^{2} 27^{\circ}}{\cos ^{2} 17^{\circ}+\sin ^{2} 17^{\circ}}=1\)

[ ∵ cos^{2} A + sin^{2} A = 1]

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

∵ sin 25° = sin (90° – 65°) = cos 65° [ ∵ sin (90° – A) = cos A]

cos 25° = cos (90° – 65°) = sin 65° [ ∵ cos (90° – A) = sin A]

∴ sin 25° cos 65° + cos 25° sin 65°

= cos 65° cos 65° + sin 65° sin 65°

= (cos 65°)^{2} + (sin 65°)^{2}

= cos^{2} 65° + sin^{2} 65° = 1 [∵ cos^{2} A + sin^{2} A = 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 +tanθ + secθ) (1 + cotθ – cosec0) =

(A) 0

(B) 1

(C) 2

(D) – 1

(iii) (sec A + tan A) (1 – sinA) =

(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) (B): Since, 9 sec^{2} A – 9 tan^{2} A

= 9 (sec^{2} A – tan^{2} A) = 9 (1) = 9 [∵ sec^{2} A – tan^{2} A = 1]

(ii) (C): Here,

Question 5.

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

Solution:

In this post, we will share NCERT Class 10th Maths Book Solutions Chapter 8 Introduction to Trigonometry Ex 8.4. These solutions are based on new NCERT Syllabus.

## NCERT 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:

Question 2.

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

Solution:

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

∵ sin 63° = sin (90° – 27°) = cos 27°

⇒ sin^{2} 63° = cos^{2} 27°

cos^{2} 73° = cos^{2} (90° – 17°) = sin^{2} 17°

∴ \(\frac{\sin ^{2} 63^{\circ}+\sin ^{2} 27^{\circ}}{\cos ^{2} 17^{\circ}+\cos ^{2} 73^{\circ}}=\frac{\cos ^{2} 27^{\circ}+\sin ^{2} 27^{\circ}}{\cos ^{2} 17^{\circ}+\sin ^{2} 17^{\circ}}=1\)

[ ∵ cos^{2} A + sin^{2} A = 1]

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

∵ sin 25° = sin (90° – 65°) = cos 65° [ ∵ sin (90° – A) = cos A]

cos 25° = cos (90° – 65°) = sin 65° [ ∵ cos (90° – A) = sin A]

∴ sin 25° cos 65° + cos 25° sin 65°

= cos 65° cos 65° + sin 65° sin 65°

= (cos 65°)^{2} + (sin 65°)^{2}

= cos^{2} 65° + sin^{2} 65° = 1 [∵ cos^{2} A + sin^{2} A = 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 +tanθ + secθ) (1 + cotθ – cosec0) =

(A) 0

(B) 1

(C) 2

(D) – 1

(iii) (sec A + tan A) (1 – sinA) =

(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) (B): Since, 9 sec^{2} A – 9 tan^{2} A

= 9 (sec^{2} A – tan^{2} A) = 9 (1) = 9 [∵ sec^{2} A – tan^{2} A = 1]

(ii) (C): Here,

Question 5.

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

Solution: