﻿ Class 10 NCERT Math Solution
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TOPICS
Exercise - 8.2

Question-1 :-  Evaluate: (i) sin 60° cos 30° + sin 30° cos 60°

Solution :-
```  sin 60° cos 30° + sin 30° cos 60°
= √3/2 x √3/2 + 1/2 x 1/2
= 3/4 + 1/4
= (3 + 1)/4
= 4/4
= 1
```

(ii)  2 tan² 45° + cos² 30° – sin² 60°

Solution :-
```  2 tan² 45° + cos² 30° – sin² 60°
= 2 x 1 + (√3/2)² - (√3/2)²
= 2 + 3/4 - 3/4
= 2
```

(iii) Solution :-
``` ```

(iv) Solution :-
``` ```

(v) Solution :-
``` ```

Question-2 :-  Choose the correct option and justify your choice : (i) (A) sin 60°  (B) cos 60°  (C) tan 60°  (D) sin 30°

Solution :-
``` Therefore, sin 60° = √3/2.
So, Option A is correct Answer.
```

(ii) (A) tan 90°  (B) 1  (C) sin 45°  (D) 0

Solution :-
``` So, Option D is correct Answer.
```

(iii)  sin 2A = 2 sin A is true when A =
(A) 0°   (B) 30°  (C) 45°  (D) 60°

Solution :-
```  sin 2A = sin 0° = 0
2 sin A = 2 sin 0° = 2 x 0 = 0
So, Option A is correct Answer.
```

(iv) (A) cos 60°  (B) sin 60°  (C) tan 60°  (D) sin 30°

Solution :-
``` Therefore, tan 60° = √3/2.
So, Option C is correct Answer.
```

Question-3 :-  If tan (A + B) = √3 and tan (A – B) = 1/√3; 0° < A + B ≤ 90°; A > B, find A and B.

Solution :-
```  Since, tan (A + B) = √3,
tan (A + B) = tan 60°
Therefore, A + B = 60° .....(1)
Also, since tan (A – B) = 1/√3,
tan (A – B) = tan 30°
Therefore, A - B = 30° ......(2)
Solving (1) and (2),
A + B + A - B = 60° + 30°
2A = 90°
A = 90°/2
A = 45°
Put in (1) equation
A - B = 30°
45° - B = 30°
-B = 30° - 45°
-B = -15°
B = 15°
we get : A = 45° and B = 15°.
```

Question-4 :-  State whether the following are true or false. Justify your answer.
(i) sin (A + B) = sin A + sin B.
(ii) The value of sin θ increases as θ increases.
(iii) The value of cos θ increases as θ increases.
(iv) sin θ = cos θ for all values of θ.
(v) cot A is not defined for A = 0°.

Solution :-
```(i) sin (A + B) = sin A + sin B.
Let A = 30° and B = 60°
sin(30° + 60°) = sin 90° = 1
sin 30° + sin 60° = 1/2 + √3/2 = (1 + √3)/2
So, statement is not equal and it is false statement.
```
```(ii) The value of sin θ increases as θ increases.
sin 0° = 0
sin 30° = 1/2 = 0.5
sin 45° = 1/√2 = 0.7
sin θ increases as θ increases
So, this statement is true.
```
```(iii) The value of cos θ increases as θ increases.
cos 0° = 1
cos 30° = √3/2 = 0.8
cos 45° = 1/√2 = 0.7
Therefore, cos θ decreases as θ increases.
So, this statement is false.
```
```(iv) sin θ = cos θ for all values of θ.
sin 0° = 0, cos 0° = 1
sin 30° = 1/2, cos 30° = √3/2
sin 45° = 1/√2, cos 45° = 1/√2
sin 60° = √3/2, cos 60° = 1/2
So, this statement is false.
```
```(v) cot A is not defined for A = 0°.
cot A = cos A/sin A
cot 0° = cos 0°/sin 0°
cot 0° = 1/0 = not defined
So, this statement is true.
```
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