﻿ Class 11 NCERT Math Solution
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TOPICS
Exercise - 3.2

Question-1 :-  Find the values of other five trigonometric functions, cos x = –1/2, x lies in third quadrant.

Solution :-
```  Given that cos x = –1/2
(i) sec x = 1/cos x = -2/1 = -2
(ii) sin x = √1 - cos²x
= √1 - (-1/2)²x
= √1 - 1/4
= √3/4
= ±√3/2
Since x lies in third quadrant, sin x will be negative. Therefore
sin x = -√3/2
(iii) cosec x = 1/sin x = -2/√3
(iv) tan x = sin x/cos x
= -√3/2 ÷ (-1)/2
= -√3/2 x 2/(-1)
= √3
(v) cot x = 1/tan x = 1/√3
```

Question-2 :-  Find the values of other five trigonometric functions, sin x = 3/5, x lies in second quadrant.

Solution :-
```  Given that sin x = 3/5
(i) cosec x = 1/sin x = 5/3
(ii) cos x = √1 - sin²x
= √1 - (3/5)²x
= √1 - 9/25
= √16/25
= ±4/5
Since x lies in second quadrant, cos x will be negative. Therefore
cos x = -4/5
(iii) sec x = 1/cos x = -5/4
(iv) tan x = sin x/cos x
= 3/5 ÷ (-4)/5
= 3/5 x 5/(-4)
= -3/4
(v) cot x = 1/tan x = -4/3
```

Question-3 :-  Find the values of other five trigonometric functions, cot x = 3/4, x lies in third quadrant.

Solution :-
```  Given that cot x = 3/4
(i) tan x = 1/cot x = 4/3
(ii) sec x = √1 + tan²x
= √1 + (4/3)²x
= √1 + 16/9
= √25/9
= ±5/3
Since x lies in third quadrant, sec x will be negative. Therefore
sec x = -5/3
(iii) cos x = 1/sec x = -3/5
(iv) sin x = √1 - cos²x
= √1 - (-3/5)²x
= √1 - 9/25
= √16/25
= ±4/5
Since x lies in third quadrant, sin x will be negaitive. Therefore
sin x = -4/5
(v) cosec x = 1/sin x = -5/4
```

Question-4 :-  Find the values of other five trigonometric functions, sec x = 13/5, x lies in fourth quadrant.

Solution :-
```  Given that sec x = 13/5
(i) cos x = 1/sec x = 5/13
(ii) sin x = √1 - cos²x
= √1 - (5/13)²x
= √1 - 25/169
= √144/169
= ±12/13
Since x lies in third quadrant, sin x will be negative. Therefore
sin x = -12/13
(iii) cosec x = 1/sin x = -13/12
(iv) tan x = sin x/cos x
= -12/13 ÷ 5/13
= -12/13 x 13/5
= -12/5
(v) cot x = 1/tan x = -5/12
```

Question-5 :-  Find the values of other five trigonometric functions, tan x = -5/12, x lies in second quadrant.

Solution :-
```  Given that tan x = -5/12
(i) cot x = 1/tan x = -12/5
(ii) sec x = √1 + tan²x
= √1 + (-5/12)²x
= √1 + 25/144
= √169/144
= ±13/12
Since x lies in second quadrant, sec x will be negative. Therefore
sec x = -13/12
(iii) cos x = 1/sec x = -12/13
(iv) sin x = √1 - cos²x
= √1 - (-12/13)²x
= √1 - 144/169
= √25/169
= ±5/13
Since x lies in second quadrant, sin x will be positive. Therefore
sin x = 5/13
(v) cosec x = 1/sin x = 13/5
```

Question-6 :-  Find the values of the trigonometric function sin 765°.

Solution :-
```  We know that values of sin x repeats after an interval of 2π or 360°.
Therefore,
sin (765°) = sin ( 2 x 360° + 45°)
= sin (45°)
= 1/√2
```

Question-7 :-  Find the values of the trigonometric function cosec (– 1410°).

Solution :-
```  We know that values of cosec x repeats after an interval of 2π or 360°.
Therefore,
cosec (-1410°) = cosec ( 4 x 360° - 1410°)
= cosec ( 1440° - 1410°)
= cosec (30°)
= 2
```

Question-8 :-  Find the values of the trigonometric function tan 19π/3.

Solution :-
```  We know that values of tan x repeats after an interval of 2π.
Therefore,
tan 19π/3 = tan(6π + π/3)
= tan π/3
= √3
```

Question-9 :-  Find the values of the trigonometric function sin (–11π/3).

Solution :-
```  We know that values of sin x repeats after an interval of 2π.
Therefore,
sin (-11π/3) = sin(4π - 11π/3)
= sin π/3
= √3/2
```

Question-10 :-  Find the values of the trigonometric function cot (–15π/4).

Solution :-
```  We know that values of cot x repeats after an interval of 2π.
Therefore,
cot (–15π/4) = cot(4π - 15π/4)
= cot π/4
= 1
```
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