GSEB Solutions Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.4

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

Gujarat Board Textbook Solutions Class 10 Maths 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. (CBSE 2012)
Solution:

tan A = \(\frac {1}{cot A}\)

Question 2.
Write all the other trigonometric ratios of ∠A in terms of sec A.
Solution:
sin A = \(\sqrt{\sin ^{2} A}\) = \(\sqrt{1-\cos ^{2} A}\)

cos A = \(\frac {1}{sec A}\)
tan A = \(\sqrt{\tan ^{2} A}\) = \(\sqrt{\sec ^{2} A-1}\) ……..(2)

Question 3.
(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) \(\frac{\sin ^{2} 63^{\circ}+\sin ^{2} 27^{\circ}}{\cos ^{2} 17^{\circ}+\cos ^{2} \cdot 73^{\circ}}\)
= \(\frac{\sin ^{2}\left(90^{\circ}-27^{\circ}\right)+\sin ^{2} 27^{\circ}}{\cos ^{2} 17+\cos ^{2}\left(90^{\circ}-17^{\circ}\right)}\)
= \(\frac{\cos ^{2} 27^{\circ}+\sin ^{2} 27^{\circ}}{\cos ^{2} 17^{\circ}+\sin ^{2} 17^{\circ}}=\frac{1}{1}\)
= 1

(ii) sin 25° cog 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

Question 4.
Choose the correct option. Justify your choice.
(i) 9 sec² θ – 9 tan² θ =
(a) 1
(b) 9
(c) 8
(d) 0

(ii) (1 + tan θ + sec θ) (1 + cot θ – cosec θ)=
(a) –
(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) \(\frac{1+\tan ^{2} A}{1+\cot ^{2} A}=1\) = 1
(a) sec 2A
(b) -1
(c) cot 2A
(d) tan 2A
Solution:
(i) 9 sec 2A – 9 tan 2A
= 9 (sec 2A – tan 2A) = 9 x 1 = 9

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

Hence, correct answer is (c) 2.

(iii) (sec A + tan A) (1 – sin A)
= \(\left[\frac{1}{\cos A}+\frac{\sin A}{\cos A}\right]\) [ 1 – sin A]
= \(\left[\frac{1+\sin \mathrm{A}}{\cos \mathrm{A}}\right]\) [ 1 – sin A]
= \(\frac{1-\sin ^{2} A}{\cos A}\)
= \(\frac{\cos ^{2} A}{\cos A}\) = cos A
Hence, correct answer is (d) cos A.

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

= tan² A
Hence, the correct answer is (d) tan² A.

Question 5.
Prove that the following identities, where the angles involved are acute angles for which the expression are defined.
(i) (cosec θ – cot θ)² = \(\frac{1-\cos \theta}{1+\cot \theta}\)
(ii) \(\frac{\cos A}{1+\sin A}\) + \(\frac{1+\sin A}{\cos A}\) = 2 sec A
(iii) \(\frac{\cos A}{1+\sin A}\) + \(\frac{\cos A}{1+\sin A}\) = 1 + sec θ cosec θ
(Hint: write the expression in terms of sin θ and cos θ)
(iv) \(\frac{\cos A}{1+\sin A}\) + \(\frac{\cos A}{1+\sin A}\)
(Hint: simplify LHS and RHS separately)
(v) \(\frac{\cos A-\sin A+1}{\cos A+\sin A-1}\) = cosec A + cot A
using identitify cosec² = 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 + (cot A + sec A)²
= 7 + tan² A + cot² A (CBSE 2012)
(ix) (cosec A – sin A)(sec A – cos A)
= \(\frac{1}{\tan A+\cot A}\)
(Hint: simplify LHS and RHS separately)
(x) \(\left[\frac{1+\tan ^{2} A}{1+\cot ^{2} A}\right]\) = \(\left[\frac{1-\tan A}{1-\cot A}\right]^{2}\)
Solution:
(i) We have
(cosec θ – cot θ)² = \(\frac{1-\cos \theta}{1+\cot \theta}\)
LHS = (cosec θ – cot θ)²

(ii) We have

(viii) (sin A + cosec A)² + (cot A + see A)²
= 7 + tan² A + cot² A
LHS = (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
= sin²A + cos²A + 1 + cot² A + 2 sin A x
\(\frac{1}{\sin A}\)+ 1+ tan² A + 2cos A x \(\frac{1}{\cos A}\)
= 1 + 1 + cot²A + 2 + 1 + tan²A + 2
= 7 + tan²A + cot²A = RHS

(ix) (cosee A – sin A) (sec A – cos A)
= \(\frac{1}{\tan A+\cot A}\)
LHS = (cosee A – sin A) (sec A – cos A)
= \(\left(\frac{1}{\sin A}-\sin A\right)\) \(\left(\frac{1}{\cos A}-\cos A\right)\)
= \(\left(\frac{1-\sin ^{2} A}{\sin A}\right)\) \(\left(\frac{1-\cos ^{2} A}{\cos A}\right)\)