﻿ Class 10 NCERT Math Solution
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TOPICS
Exercise - 8.3

Question-1 :-  Evaluate: (i) sin 18°/cos 72°

Solution :-
```  We know : cos A = sin (90° – A)
So, cos 72° = sin (90° – 72°) = sin 18°
i.e., sin 18°/cos 72°  = sin 18°/sin 18° = 1
```

(ii)  tan 26°/cot 64°

Solution :-
```  We know : cot A = tan (90° – A)
So, cot 64° = tan (90° – 64°) = tan 26°
i.e., tan 26°/cot 64°  = tan 26°/tan 26°  = 1
```

(iii)  cos 48° – sin 42°

Solution :-
```  cos 48° – sin 42°
= cos(90° - 42°) - sin 42°
= sin 42° - sin 42°
= 0
```

(iv)  cosec 31° – sec 59°

Solution :-
```  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

Solution :-
```(i) tan 48° tan 23° tan 42° tan 67° = 1
L.H.S
tan 48°. tan 23° . tan 42° . tan 67°
= tan 48° . tan 42° . tan 23° . tan 67°
= tan 48° . tan(90° - 48°) . tan 23° . tan(90° - 23°)
= tan 48° . cot 48° . tan 23° . cot 23°
= tan 48° . 1/tan 48° . tan 23° . 1/tan 23°
= 1
```
```(ii) cos 38° cos 52° – sin 38° sin 52° = 0
L.H.S
cos 38° . cos 52° – sin 38° . sin 52°
= cos(90° - 52°) . cos 52° - sin(90° - 52°) . sin 52°
= sin 52° . cos 52° - cos 52° . sin 52°
= 0
```

Question-3 :-  If tan 2A = cot (A – 18°), where 2A is an acute angle, find the value of A.

Solution :-
```  We are given that tan 2A = cot (A – 18°)..... (1)
Since tan 2A = cot (90° – 2A),
we can write (1) as cot (90° – 2A) = cot (A – 18°)
Since 90° – 2A and A – 18° are both acute angles,
Therefore,
90° – 2A = A – 18°
A + 2A = 90° + 18°
3A = 108°
A = 108°/3
which gives A = 36°
```

Question-4 :-  If tan A = cot B, prove that A + B = 90°.

Solution :-
```  Given :
tan A = cot B
we Know that, tan(90° - B) = 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.

Solution :-
```  We are given that sec 4A = cosec (A – 20°)..... (1)
Since sec 3A = cosec (90° – 4A),
we can write (1) as cosec (90° – 4A) = cosec (A – 20°)
Since 90° – 4A and A – 20° are both acute angles,
Therefore,
90° – 4A = A – 20°
A + 4A = 90° + 20°
5A = 110°
A = 110°/5
which gives A = 22°
```

Question-6 :-  If A, B and C are interior angles of a triangle ABC, then show that sin (B+C)/2 = cos A/2.

Solution :-
```  We know that sum of angle a triangle = 180°
A  + B + C = 180
B + C = 180 - A
(B + C)/2 = 90 - A/2
sin (B + C)/2 = sin (90 - A/2)
sin (B + C)/2 = cos A/2
```

Question-7 :-  Express sin 67° + cos 75° in terms of trigonometric ratios of angles between 0° and 45°.

Solution :-
```  sin 67° + cos 75°
= sin (90° – 23°) + cos (90° – 15°)
= cos 23° + sin 15°
```
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