# NCERT solution class 10 chapter 8 Introduction to Trigonometry exercise 8.3 mathematics

## EXERCISE 8.3

#### Question 1:

Evaluate

(I)

(II)

(III) cos 48° − sin 42°

(IV)cosec 31° − sec 59°

(I)

(II)

(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

(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°

= (cot 42° tan 42°) (cot 67° tan 67°)

= (1) (1)

= 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.

Given that,

tan 2A = cot (A− 18°)

cot (90° − 2A) = cot (A −18°)

90° − 2A = A− 18°

108° = 3A

A = 36°

#### Question 4:

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

Given that,

tan A = cot B

tan A = tan (90° − B)

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.

Given that,

sec 4A = cosec (A − 20°)

cosec (90° − 4A) = cosec (A − 20°)

90° − 4A= A− 20°

110° = 5A

A = 22°

#### Question 6:

If A, Band C are interior angles of a triangle ABC then show that

We know that for a triangle ABC,

∠ A + ∠B + ∠C = 180°

∠B + ∠C= 180° − ∠A

#### Question 7:

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