HBSE 10th Class Maths Solutions Chapter 8 Introduction to Trigonometry Ex 8.4

Haryana State Board HBSE 10th Class Maths Solutions Chapter 8 Introduction to Trigonometry Ex 8.4 Textbook Exercise Questions and Answers.

Haryana Board 10th Class 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. Write the trigonometric ratio of sin A in terms of cot A.
Solution :
We have,
cosec² A = 1 + cot² A

Hence, sin A = \(\frac{1}{\sqrt{1+\cot ^2 A}}\),
sec A = \(\frac{\sqrt{\cot ^2 \mathrm{~A}+1}}{\cot \mathrm{A}}\) and
tan A = \(\frac{1}{\cot A}\).

Question 2.
Write all the other trigonometric ratios of A in terms of sec A. Express the trignometric ratio of tan A in terms of sec A.
SoJution :
We have,
sin² A + cos² A = 1
sin² A = 1 – cos² A
sin² A = 1 – \(\frac{1}{\sec ^2 A}\)
sin² A = \(\frac{\sec ^2 A-1}{\sec ^2 A}\)
sin A = \(\sqrt{\frac{\sec ^2 \mathrm{~A}-1}{\sec ^2 \mathrm{~A}}}\)
sin A = \(\frac{\sqrt{\sec ^2 \mathrm{~A}-1}}{\sec \mathrm{A}}\)
∵ cos A = \(\frac{1}{\sec \bar{A}}\)

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) \(\frac{\sin ^2 63^{\circ}+\sin ^2 27^{\circ}}{\cos ^2 17^{\circ}+\cos ^2 73^{\circ}}\)
= \(\frac{\sin ^2\left(90^{\circ}-27^{\circ}\right)+\sin ^2 27^{\circ}}{\cos ^2\left(90^{\circ}-73^{\circ}\right)+\cos ^2 73^{\circ}}\)
= \(\frac{\cos ^2 27^{\circ}+\sin ^2 27^{\circ}}{\sin ^2 73^{\circ}+\cos ^2 73^{\circ}}\)
= \(\frac{1}{1}\) = 1 [∵ sin² θ + cos² θ = 1]
Hence, \(\frac{\sin ^2 63^{\circ}+\sin ^2 27^{\circ}}{\cos ^2 17^{\circ}+\cos ^2 73^{\circ}}\) = 1.

(ii) sin 25° cos 65° + cos 25° sin 65°
= sin 25° cos (90° – 25°) + cos 25° sin (90° – 25°)
= sin 25° sin 25° + cos 25° cos 25°
= sin² 25° + cos² 25°
= 1 [∵ sin² θ + cos² θ = 1]
Hence, sin 25° cos 65° + cos 25° sin 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) – 1

(iii) (sec A + tan A) (1 – sin A) =
(A) sec A
(B) sin A
(C) cosec A
(D) cos A

(iv) \(\frac{1+\tan ^2 \mathrm{~A}}{1+\cot ^2 \mathrm{~A}}\) =
(A) sec² A
(B) – 1
(C) cot² A
(D) tan² A

Solution :
(i) (B)
Because; 9 sec² A – 9 tan² A = 9 (sec² A – tan² A)
= 9 × 1
[∵ sec² A – tan² A = 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)
= \(\left(\frac{1}{\cos A}+\frac{\sin A}{\cos A}\right)\) (1 – sin A)
= \(\frac{(1+\sin \mathrm{A})(1-\sin \mathrm{A})}{\cos \mathrm{A}}\)
= \(\frac{1-\sin ^2 \mathrm{~A}}{\cos \mathrm{A}}\)
= \(\frac{\cos ^2 A}{\cos \mathrm{A}}\) = cos A
[∵ 1 – sin² A = cos² A]

(iv) (D) Because;

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 cosec θ
(iv) \(\frac{1+\sec \mathrm{A}}{\sec \mathrm{A}}=\frac{\sin ^2 \mathrm{~A}}{1-\cos \mathrm{A}}\)
(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) = \(\frac{1}{\tan A+\cot A}\)
(x) \(\left(\frac{1+\tan ^2 \mathrm{~A}}{1+\cot ^2 \mathrm{~A}}\right)=\left(\frac{1-\tan \mathrm{A}}{1-\cot \mathrm{A}}\right)^2\) = tan² A
Solution:
(i) We have,
L.H.S = (cosec θ – cot θ)²

L.H.S. = R.H.S.
Hence proved.

(ii) We have,

L.H.S. = R.H.S.
Hence proved.

(iii) We have,

= cosec θ sec θ + 1
= 1 + sec θ cosec θ = R.H.S
∴ L.H.S. = R.H.S.

(iv) We have,

[∵ 1 – cos² A = sin² A]
∴ L.H.S. = R.H.S.

(v) We have,
L.H.S = \(\frac{\cos A-\sin A+1}{\cos A+\sin A-1}\)
= \(\frac{\frac{\cos A}{\sin A}-\frac{\sin A}{\sin A}+\frac{1}{\sin A}}{\frac{\cos A}{\sin A}+\frac{\sin A}{\sin A}-\frac{1}{\sin A}}\)
[Dividing numerator and denominator by sin A]

= cosec A + cot A = R.H.S.
∴ L.H.S. = R.H.S.
Hence proved.

(vi) We have,

= sec A + tan A
= R.H.S.
∴ L.H.S. = R.H.S.

(vii) We have,

= tan θ = R.H.S
∴ L.H.S. = R.H.S.

(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 + cos² A) + (cosec² A) + (sec² A) + 2 sin A × \(\frac{1}{\sin A}\) + 2 cos A \(\frac{1}{\cos A}\)
= 1 + 1 + cot² A + 1 + tan² A + 2 + 2
= 7 + tan² A + cot² A
= R.H.S
∴ L.H.S. = R.H.S.
Hence Proved.

(ix) We have,

= sin A cos A
∴ L.H.S. = R.H.S.
Hence Proved.

(x) We have,

= tan² A
∴ L.H.S. = R.H.S.
Hence Proved.