﻿ Class 10 NCERT Math Solution
﻿
TOPICS
Exercise - 8.4

Question-1 :-  Express the trigonometric ratios sin A, sec A and tan A in terms of cot A.

Solution :-
tan A = 1/cot A
sec A =  √1 + tan² A
=  √1 + 1/cot² A
=  √(cot² A + 1)/cot A
sin A = 1/cosec A = 1/√1 + cot² A

Question-2 :-  Write all the other trigonometric ratios of ∠ A in terms of sec A.

Solution :-
cos A = 1/sec A
tan A = √sec² A - 1
cot A = 1/tan A = 1/√sec² A - 1
sin A = √1 - cos² A
= √1 - 1/sec² A
= √(sec² A - 1)/sec² A
= √(sec² A - 1)/sec A
cosec A = 1/sin A = sec A/√sec² A - 1

Question-3 :-  Evaluate : (i)
(ii) sin 25° cos 65° + cos 25° sin 65°

Solution :-
(i)
(ii) sin 25° cos 65° + cos 25° sin 65°
=  sin (90° - 65°) cos 65° + cos (90° - 65°) sin 65°
=  cos 65° . cos 65° + sin 65° . sin 65°
=  cos² 65° + sin² 65°     [sin² A + cos² A = 1]
=  1

Question-4 :-  Choose the correct option. Justify your choice.
(i) 9 sec² A – 9 tan² A =
(A) 1  (B) 9  (C) 8  (D) 0

Solution :-
9 sec² A – 9 tan² A
= 9(1 + tan² A) - 9 tan² A
= 9 + 9 tan² A - 9 tan² A
= 9
So, Option B is correct Answer.

(ii)  (1 + tan θ + sec θ) (1 + cot θ – cosec θ) =
(A) 0  (B) 1  (C) 2  (D) –1

Solution :-

So, Option C is correct Answer.

(iii)  (sec A + tan A) (1 – sin A) =
(A) sec A  (B) sin A  (C) cosec A  (D) cos A

Solution :-

So, Option D is correct Answer.

(iv)
(A) sec² A  (B) –1  (C) cot² A  (D) tan² A

Solution :-

So, Option D is correct Answer.

Question-5 :-  Prove the following identities, where the angles involved are acute angles for which the expressions are defined.
(i) Solution :-

(ii)

Solution :-

(iii)

Solution :-

(iv)

Solution :-

(v)

Solution :-

(vi)

Solution :-

(vii)

Solution :-

(viii)

Solution :-

(xi)

Solution :-

(x)

Solution :-

CLASSES