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°.

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