﻿ Class 10 NCERT Math Solution
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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 :-

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(ii) Solution :-
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(iii) Solution :-
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(iv) Solution :-
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(v) Solution :-
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(vi) Solution :-
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(vii) Solution :-
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(viii) Solution :-
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(xi) Solution :-
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(x) Solution :-
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