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 âˆ 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 25Â° cos 65Â° + cos 25Â° sin 65Â°
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 25Â° cog 65Â° + cos 25Â° sin 65Â°
= sin 25Â° cos (90Â° – 25Â°) + cos 25Â° sin (90Â° – 25Â°)
= sin 25Â° sin 25Â° + cos 25Â° cos 25Â°
= sin2 25Â° + cos2 25
= 1

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

(ii) (1 + tan Î¸ + sec Î¸) (1 + cot Î¸ – cosec Î¸)=
(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 Î¸ + sec Î¸) (1 + cot Î¸ – cosec Î¸)

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 Î¸ – cot Î¸)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 Î¸ cosec Î¸
(Hint: write the expression in terms of sin Î¸ and cos Î¸)
(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 Î¸
(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 Î¸ – cot Î¸)2 = $$\frac{1-\cos \theta}{1+\cot \theta}$$
LHS = (cosec Î¸ – cot Î¸)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)
= $$\left(\frac{1}{\sin A}-\sin A\right)$$ $$\left(\frac{1}{\cos A}-\cos A\right)$$
= $$\left(\frac{1-\sin ^{2} A}{\sin A}\right)$$ $$\left(\frac{1-\cos ^{2} A}{\cos A}\right)$$