﻿ Class 11 NCERT Math Solution
﻿
TOPICS
Exercise - 3.3

Question-1 :-  Proove that: sin² π/6 + cos² π/3 - tan² π/4 = -1/2

Solution :-
```  L.H.S
sin² π/6 + cos² π/3 - tan² π/4
= (sin π/6)² + (cos π/3)² - (tan π/4)²
= (1/2)² + (1/2)² - (1)²
= 1/4 + 1/4 - 1
= 2/4 - 1
= 1/2 - 1
= -1/2
= R.H.S
```

Question-2 :-  Proove that: 2sin² π/6 + cosec² 7π/6 . cos² π/3 = 3/2

Solution :-
```  L.H.S
2sin² π/6 + cosec² 7π/6 . cos² π/3
= 2(sin π/6)² + (cosec 7π/6)² . (cos π/3)²
= 2 x (1/2)² + [cosec(π + π/6)]² . (1/2)²
= 2 x 1/4 + (-cosec π/6)² . 1/4
= 1/2 + (-2)² . 1/4
= 1/2 + 4 x 1/4
= 1/2 + 1
= 3/2
= R.H.S
```

Question-3 :-  Proove that: cot² π/6 + cosec 5π/6 + 3 tan² π/6 = 6

Solution :-
```  L.H.S
cot² π/6 + cosec 5π/6 + 3 tan² π/6
= (cot π/6)² + cosec 5π/6 + 3 (tan π/6)²
= (√3)² + cosec(π - π/6) + 3 (1/√3)²
= 3 + cosec π/6 + 3 x 1/3
= 3 + 2 + 1
= 6
= R.H.S
```

Question-4 :-  Proove that: 2sin² 3π/4 + 2cos² π/4 + 2sec² π/3 = 10

Solution :-
```  L.H.S
2sin² 3π/4 + 2cos² π/4 + 2sec² π/3
= 2(sin 3π/4)² + 2(cos π/4)² + 2(sec π/3)²
= 2[sin(π - π/4)]² + 2 x (1/√2)² + 2 x (2)²
= 2(sin π/4)² + 2 x 1/2 + 2 x 4
= 2 x (1/√2)² + 1 + 8
= 2 x 1/2 + 1 + 8
= 1 + 1 + 8
= 10
= R.H.S
```

Question-5 :-  Find the value of : (i) sin 75°   (ii) tan 15°.

Solution :-
```(i) sin 75°
= sin(45° + 30°)
= sin 45° . cos 30° + cos 45° . sin 30°   [sin(a + b) = sin a . cos b + cos a . sin b]
= 1/√2 . √3/2 + 1/√2 . 1/2
= √3/2√2 + 1/2√2
= (√3 + 1)/2√2

(ii) tan 15°
= tan(45° - 30°)
= (tan 45° - tan 30°)/(1 + tan 45° . tan 30°)  [tan(a - b) = (tan a - tan b)/(1 + tan a . tan b)]
= (1 - 1/√3)/(1 + 1/√3)
= (√3 - 1)/(√3 + 1)
By Rationalizing of denominator
= (√3 - 1)/(√3 + 1) x (√3 - 1)/(√3 - 1)
= (√3 - 1)²/(√3² - 1²)
= (3 + 1 - 2√3)/(3 - 1)
= 2(2 - √3)/2
= 2 - √3
```

Question-6 :-  Proove that: cos(π/4 - x) . cos(π/4 - y) - sin(π/4 - x) . sin(π/4 - y) = sin(x + y)

Solution :-
```  L.H.S
cos(π/4 - x) . cos(π/4 - y) - sin(π/4 - x) . sin(π/4 - y)
= cos(π/4 - x + π/4 - y)      [cos(a + b) = cos a . cos b - sin a . sin b]
= cos(2π/4 - x - y)
= cos[π/2 - (x + y)]
= sin(x + y)
```

Question-7 :-  Proove that:

Solution :-
```        L.H.S

```

Question-8 :-  Proove that:

Solution :-
```        L.H.S

```

Question-9 :-  Prove that:

Solution :-
```        L.H.S

```

Question-10 :-  Proove that: sin (n + 1)x . sin (n + 2)x + cos (n + 1)x . cos (n + 2)x = cos x

Solution :-
```  L.H.S
sin (n + 1)x . sin (n + 2)x + cos (n + 1)x . cos (n + 2)x
= sin (nx + x) . sin (nx + 2x) + cos (nx + x) . cos (nx + 2x)
= cos(nx + 2x - nx - x)      [cos(a - b) = cos a . cos b + sin a . sin b]
= cos x
```

Question-11 :-  Proove that: cos(3π/4 + x) - cos(3π/4 - x) = -√2 sin x

Solution :-
```        L.H.S

```

Question-12 :-  Probe that: sin² 6x – sin² 4x = sin 2x sin 10x

Solution :-
```  L.H.S
sin² 6x – sin² 4x
By using properties,
[sin a + sin b = 2 sin (a + b)/2 . cos (a - b)/2]
[sin a - sin b = 2 cos (a + b)/2 . sin (a - b)/2]

= (sin 6x + sin 4x)(sin 6x – sin 4x)
= (2 sin 10x/2 . cos 2x/2)(2 cos 10x/2 . sin 2x/2)
= (2 sin 5x . cos x)(2 cos 5x . sin x)
= (2 sin 5x . cos 5x)(2 sin x . cos x)
= sin 10x . sin 2x
= R.H.S
```

Question-13 :-  Proove that: cos² 2x – cos² 6x = sin 4x sin 8x

Solution :-
```  L.H.S
cos² 2x – cos² 6x
By using properties,
[cos a + cos b = 2 cos (a + b)/2 . cos (a - b)/2]
[cos a - cos b = -2 sin (a + b)/2 . sin (a - b)/2]

= (cos 2x + cos 6x)(cos 2x – cos 6x)
= [2 cos 8x/2 . cos (-4x/2)][-2 sin 8x/2 . sin (-4x/2)]
= (2 cos 4x . cos 2x)(2 sin 4x . sin 2x)                        [cos(-x) = x and sin(-x) = -x]
= (2 cos 4x . sin 4x)(2 sin 2x . cos 2x)
= sin 8x . sin 4x
= R.H.S
```

Question-14 :-  Proove that: sin 2x + 2 sin 4x + sin 6x = 4 cos² x sin 4x

Solution :-
```  L.H.S
sin 2x + 2 sin 4x + sin 6x
= (sin 2x + sin 6x) + 2 sin 4x               [sin a + sin b = 2 sin (a + b)/2 . cos (a - b)/2]
= 2 sin 8x/2 . cos (-4x/2) + 2 sin 4x
= 2 sin 4x . cos 2x + 2 sin 4x
= 2 sin 4x (cos 2x + 1)                      [cos 2x = 2 cos²x - 1]
= 2 sin 4x (2 cos²x - 1 + 1)
= 4 sin 4x cos²x
= R.H.S
```

Question-15 :- Prove That: cot 4x (sin 5x + sin 3x) = cot x (sin 5x – sin 3x)

Solution :-
```  L.H.S
cot 4x (sin 5x + sin 3x)
= cos 4x/sin 4x (sin 5x + sin 3x)
= cos 4x/sin 4x (2 sin 8x/2 . cos 2x/2)
= cos 4x/sin 4x (2 sin 4x . cos x)
= 2 cos 4x cos x

R.H.S
cot x (sin 5x – sin 3x)
= cos x/sin x (sin 5x - sin 3x)
= cos x/sin x (2 cos 2x/2 . sin 2x/2)
= cos x/sin x (2 cos 4x . sin x)
= 2 cos 4x cos x

L.H.S = R.H.S
```

Question-16 :-  Proove that:

Solution :-
```        L.H.S

```

Question-17 :-  Proove that:

Solution :-
```        L.H.S

```

Question-18 :- Prove That:

Solution :-
```        L.H.S

```

Question-19 :-  Proove that:

Solution :-
```        L.H.S

```

Question-20 :-  Proove that:

Solution :-
```        L.H.S

```

Question-21 :- Prove That:

Solution :-
```    L.H.S

```

Question-22 :-  Proove that: cot x . cot 2x – cot 2x . cot 3x – cot 3x . cot x = 1

Solution :-
```  L.H.S
cot x . cot 2x – cot 2x . cot 3x – cot 3x . cot x
= cot x . cot 2x –  cot 3x(cot 2x + cot x)
= cot x . cot 2x –  cot(2x + x)(cot 2x + cot x)
= cot 2x . cot x - [(cot 2x . cot x - 1)/(cot x + cot 2x)] . (cot 2x + cot x)
= (cot 2x . cot x) - (cot 2x . cot x - 1)
= cot 2x . cot x - cot 2x . cot x - 1
= 1
= R.H.S
```

Question-23 :-  Proove that:

Solution :-
```  L.H.S
Tan 4x = tan 2(2x)

```

Question-24 :- Prove That: cos 4x = 1 – 8 sin² x . cos² x

Solution :-
```  L.H.S
cos 4x = cos 2(2x)     [cos 2x = 1 - sin²x]
= 1 - 2 sin² 2x          [sin 2x = 2 sin x . cos x]
= 1 - 2(2 sin x cos x)²
= 1 - 8 sin² x cos² x
= R.H.S
```

Question-25 :-  Proove that: cos 6x = 32 cos⁶ x – 48 cos⁴ x + 18 cos² x – 1

Solution :-
```  L.H.S
cos 6x = cos 3(2x)
= 4 cos³ 2x - 3 cos 2x
= 4(2 cos² x - 1)³ - 3 (2 cos² x - 1)
= 4[(2 cos² x)³ - 1³ - 3(2 cos² x)² + 3(2 cos² x)] - 6 cos² x + 3
= 4[8 cos⁶ x - 1 - 3(4 cos⁴ x) + 6 cos² x] - 6 cos² x + 3
= 32 cos⁶ x  - 4 - 48 cos⁴ x + 24 cos² x - 6 cos² x + 3
= 32 cos⁶ x – 48 cos⁴ x + 18 cos² x – 1
= R.H.S
```
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