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

## BSEB Bihar Board Class 10th Maths Solutions Chapter 8 Introduction to Trigonometry Ex 8.3

Question 1.

Evaluate :

(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) \(\frac{\sin 18^{\circ}}{\cos 72^{\circ}}\)

= \(\frac{\sin \left(90^{\circ}-72^{\circ}\right)}{\cos 72^{\circ}}\)

= \(\frac{\cos 72^{\circ}}{\cos 72^{\circ}}\)

= 1

(ii) \(\frac{\tan 26^{\circ}}{\cot 64^{\circ}}\)

= \(\frac{\tan \left(90^{\circ}-64^{\circ}\right)}{\cot 64^{\circ}}\)

= \(\frac{\cot 64^{\circ}}{\cot 64^{\circ}}\)

= 1

(iii) cos 48° – sin 42° = cos (90° – 42°) – sin 42°

= sin 42° – sin 42° = 0

(iv) cosec 31° – sec 59° = cosec (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) tan 48° tan 23° tan 42° tan 67°

= tan (90° – 42°) tan (90° – 67°) tan 42° tan 67°

= cot 42° cot 67° tan 42° tan 67°

= \(\frac{1}{\tan 42^{\circ}}\).\(\frac{1}{\tan 67^{\circ}}\).tan 42° tan 67° = 1

(ii) cos 38° cos 52° – sin 38° sin 52°

= cos (90° – 52°) cos (90° – 38°) – sin 38° sin 52°

= sin 52° sin 38° – sin 38° sin 52° = 0

Question 3.

If tan 2A = cot (A – 18°), where 2A is an acute angle, find the value of A.

Solution:

We are given that

tan 2A = cot (A – 18°) … (1)

Since tan 2A = cot (90° – 2A), so we can write (1) as

cot (90° – 2A) = cot (A – 18°)

Since (90° – 2A) and (A – 18°) are both acute angles, therefore

90° – 2A = A – 18°

or – 2A – A = – 18° – 90° or – 3A = – 108°

or A = 36°

Question 4.

If tan A = cot B, prove that A + B = 90°.

Solution:

We are given that

tan A = cot B … (1)

Since tan A = cot (90° – A), so we can write (1)

as cot (90° – A) = cot B

Since (90° – A) and B are both acute angles, therefore

90° – A = B or A + B = 90°

Question 5.

If sec 4A = cosec (A – 20°), where 4A is an acute angle, find the value of A.

Solution:

We are given that

sec 4A = cosec (A – 20°). …(1)

Since sec 4A = cosec (90° – 4A), so we can write (1) as

cosec (90° – 4A) = cosec (A – 20°)

Since (90° – 4A) and (A – 20°) are both acute angles, therefore

90° – 4A = A – 20° or – 4A – A = – 20° – 90°

or – 5A = – 110° or A = 22°

Question 6.

If A, B and C are, interior angles of a ∆ ABC, then show, that

sin\(\left(\frac{B+C}{2}\right)\) = cos\(\frac { A }{ 2 }\)

Solution:

Since A, B and C are the interior angles of a ∆ ABC, therefore

Question 7.

Express sin 67° + cos 75° in terms of trigonometric ratios of angles between 0° and 45°.

Solution:

sin 67° + cos 75° = sin (90° – 23°) + cos (90° – 15°)

= cos 23° + sin 15°