# GSEB Solutions Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.3

Gujarat Board GSEB Solutions Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.3 Textbook Questions and Answers.

## Gujarat Board Textbook Solutions Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.3

Question 1.
(i) $$\frac{\sin 18^{\circ}}{\cos 72^{\circ}}$$
(ii) $$\frac{\tan 26^{\circ}}{\cot 64^{\circ}}$$
(iii) cos 48Â° – sin 42Â°
(iv) cosec 31Â° – sec 59Â°
Solution:
(i) $$\frac{\sin 18^{\circ}}{\cos 72^{\circ}}$$ = $$\frac{\sin 18^{\circ}}{\cos \left(90^{\circ}-18^{\circ}\right)}$$
= sin $$\frac{\sin 18^{\circ}}{\sin 18^{\circ}}$$ = 1 (cos (90Â° – Î¸) = sin Î¸)

(ii) $$\frac{\sin 18^{\circ}}{\cos 72^{\circ}}$$ = $$\frac{\tan 26^{\circ}}{\cot \left(90^{\circ}-26^{\circ}\right)}$$
= $$\frac{\tan 26^{\circ}}{\tan 26^{\circ}}$$ = 1 (cot(90Â°- Î¸) = tan Î¸)

(iii) cos 48Â° – sin 42Â°
= cos (90Â° – 42Â°) – sin 42Â°
= sin 42Â° – sin 42Â° = 0 [cos (90Â° – Î¸) = sin Î¸]

(iv) cosec 31Â° – sec 59Â°
= cos (90Â° – 59Â°) – sec 59Â°
= sec 59Â° – sec 59Â° = 0

Question 2.
Show that
(i) tan 48Â° tan 23Â° tan 42Â° tan 67Â° = 1
(ii) cos 38Â° cos 52Â° – sin 38Â° sin 52Â° = 0
Solution:
(i) LHS = tan 48Â° tan 23Â° tan 42Â° tan 67Â°
= tan 48Â° tan 42Â° tan 23Â° tan 67Â°
= tan (90Â° – 42Â°) tan 42Â° tan 23Â° tan(90Â° – 23Â°)
= cot 42Â° tan 42Â° tan 23Â° cot 23Â°
= $$\frac {1}{tan 42Â°}$$ x tan 42Â° x tan 23Â° x $$\frac {1}{tan 23Â°}$$
=1 = RHS

(ii) LHS = cos 38Â° cos 52Â° – sin 38Â° sin 52Â°
= cos 38Â° a (90Â° – 38Â°) – 38Â° sin (90Â° – 38Â°)
= cos 38Â° sin 38Â° – sin 38Â° cos 38Â° = 0 = RHS

Question 3.
If tan 2A = cot (A – 18Â°), where 2A is an acute angle, find the value of A. (CBSE 2012)
Solution:
We have
tan 2A cot (A – 18Â°)
tan 2A = tan [90Â°- (A – 18Â°)]
tan 2A = tan [90Â° – A + 18Â°]
2A = 90Â° – A + 18Â°
3A = 108Â°
A = $$\frac {108Â°}{3}$$
A = 36Â°

Question 4.
If tan A = cot B, prove that A+ B = 90Â°.
Solution:
We have
tan A = cot B
tanA = tan(90Â° – B) [cot Î¸ = tan (90Â° – Î¸)]
A = 90Â°- B
A + B = 90Â°

Question 5.
If sec 4A = cosec (A – 20Â°) where 4A is an acute angle, find the value of A.
Solution:
We have
sec 4A = cosec (A – 20Â°)
cosec (90Â° – 4A) = cosec (A – 20Â°)
90Â° – 4A = A – 20Â°
90Â° + 20Â° = 5A
5A = 110Â°
A = $$\frac {110Â°}{5}$$ = 22Â°

Question 6.
If A, B and C are interior angles of a triangle ABC, then show that
sin $$\frac {B + C}{2}$$ = cos = $$\frac {A}{2}$$
Solution:
In a triangle,
A + B + C = 180Â°
(Sum of angles of a triangle is 180Â°)
$$\frac {A}{2}$$ + $$\frac {B}{2}$$ + $$\frac {C}{2}$$ = $$\frac {180Â°}{2}$$
(Dividing by 2 on both sides)
â‡’ $$\frac {B}{2}$$ + $$\frac {C}{2}$$ = 90Â° – $$\frac {A}{2}$$
$$\frac {B + C}{2}$$ = 90Â° – $$\frac {A}{2}$$
sin ($$\frac {B + C}{2}$$) = sin ($$\left(90^{\circ}-\frac{\mathrm{A}}{2}\right)$$)
(Taking sin on both sides)
sin ($$\frac {B + C}{2}$$) = cos $$\frac {A}{2}$$

Question 7.
Express sin 67Â° Ã· cos 75Â° in terms of trigonometric ratios of angles between 0Â° and 45Â°.
Solution:
We have
sin 67Â° + cos 75Â°
= sin (90Â° – 23Â°) + cos (90Â° – 15Â°)
= cos 23Â° + sin 15Â°