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 48° – sin 42°
(iv) cosec 31° – sec 59°
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 (90° – θ) = sin θ)
(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(90°- θ) = tan θ)
(iii) cos 48° – sin 42°
= cos (90° – 42°) – sin 42°
= sin 42° – sin 42° = 0 [cos (90° – θ) = sin θ]
(iv) cosec 31° – sec 59°
= cos (90° – 59°) – sec 59°
= sec 59° – sec 59° = 0
Question 2.
Show that
(i) tan 48° tan 23° tan 42° tan 67° = 1
(ii) cos 38° cos 52° – sin 38° sin 52° = 0
Solution:
(i) LHS = tan 48° tan 23° tan 42° tan 67°
= tan 48° tan 42° tan 23° tan 67°
= tan (90° – 42°) tan 42° tan 23° tan(90° – 23°)
= cot 42° tan 42° tan 23° cot 23°
= \frac {1}{tan 42°} x tan 42° x tan 23° x \frac {1}{tan 23°}
=1 = RHS
(ii) LHS = cos 38° cos 52° – sin 38° sin 52°
= cos 38° a (90° – 38°) – 38° sin (90° – 38°)
= cos 38° sin 38° – sin 38° cos 38° = 0 = RHS
Question 3.
If tan 2A = cot (A – 18°), where 2A is an acute angle, find the value of A. (CBSE 2012)
Solution:
We have
tan 2A cot (A – 18°)
tan 2A = tan [90°- (A – 18°)]
tan 2A = tan [90° – A + 18°]
2A = 90° – A + 18°
3A = 108°
A = \frac {108°}{3}
A = 36°
Question 4.
If tan A = cot B, prove that A+ B = 90°.
Solution:
We have
tan A = cot B
tanA = tan(90° – B) [cot θ = tan (90° – θ)]
A = 90°- B
A + B = 90°
Question 5.
If sec 4A = cosec (A – 20°) where 4A is an acute angle, find the value of A.
Solution:
We have
sec 4A = cosec (A – 20°)
cosec (90° – 4A) = cosec (A – 20°)
90° – 4A = A – 20°
90° + 20° = 5A
5A = 110°
A = \frac {110°}{5} = 22°
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 = 180°
(Sum of angles of a triangle is 180°)
\frac {A}{2} + \frac {B}{2} + \frac {C}{2} = \frac {180°}{2}
(Dividing by 2 on both sides)
⇒ \frac {B}{2} + \frac {C}{2} = 90° – \frac {A}{2}
\frac {B + C}{2} = 90° – \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 67° ÷ cos 75° in terms of trigonometric ratios of angles between 0° and 45°.
Solution:
We have
sin 67° + cos 75°
= sin (90° – 23°) + cos (90° – 15°)
= cos 23° + sin 15°