﻿ Class 12 NCERT Math Solution
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TOPICS
Miscellaneous

Question-1 :-  Find the value of cos-1 (cos 13π/6).

Solution :-
```   cos-1 (cos 13π/6)
Here, 13π/6 is not in range of cos-1 i.e., [0, π].
so, break of 13π/6
= cos-1 [cos (2π + π/6)]
= cos-1 [cos π/6]
= π/6
```

Question-2 :- Find the value of tan-1 (tan 7π/6).

Solution :-
```   tan-1 (tan 7π/6)
Here, 7π/6 is not in range of cos-1 i.e., [-π/2, π/2].
so, break of 7π/6
= tan-1 [tan (π + π/6)]
= tan-1 [tan π/6]
= π/6
```

Question-3 :-  Prove that : 2sin-1 (3/5) = tan-1 (24/7).

Solution :-
```   2sin-1 (3/5) = tan-1 (24/7)
By taking L.H.S
2sin-1 (3/5)   [2sin-1 x = sin-1(2x √1 - x²)]
= sin-1 (2 x 3/5 √1 - (3/5)²)
= sin-1 (6/5 √1 - 9/25)
= sin-1 (6/5 √16/25)
= sin-1 (6/5 x 4/5)
= sin-1 (24/25)
Let sin-1 (24/25) = θ
sin θ = 24/25
cos θ = √1 - sin²θ
cos θ = √1 - (24/25)²
cos θ = √1 - 576/625
cos θ = √49/625
cos θ = 7/25
Now, tan θ = sin θ/ cos θ = 24/25 x 25/7 = 24/7
θ = tan-1 (24/7) = R.H.S
```

Question-4 :-  Prove that: sin-1 (8/17) + sin-1 (3/5) = tan-1 (77/36).

Solution :-
```   sin-1 (8/17) + sin-1 (3/5) = tan-1 (77/36)
L.H.S
sin-1 (8/17) + sin-1 (3/5)
By Using [sin-1 x + sin-1 y = sin-1(x√1 - y² + y√1 - x²)]
= sin-1 (8/17 √1 - (3/5)² + 3/5 √1 - (8/17)²)
= sin-1 (8/17 √1 - 9/25 + 3/5 √1 - 64/289)
= sin-1 (8/17 √16/25 + 3/5 √225/289)
= sin-1 (8/17 x 4/5 + 3/5 x 15/17)
= sin-1 (32/85 + 9/17)
= sin-1 [(32 + 45)/85]
= sin-1 77/85
Let sin-1 77/85 = θ
sin θ = 77/85
cos θ = √1 - sin²θ
cos θ = √1 - (77/85)²θ
cos θ = √1 - 5929/7225
cos θ = √1296/7225
cos θ = 36/85
Now tan θ = sin θ/cos θ
tan θ = 77/85 x 85/36 = 77/36
θ = tan-1 77/36
= R.H.S
```

Question-5 :- Prove that: cos-1 (4/5) + cos-1 (12/13) = cos-1 (33/65).

Solution :-
```   cos-1 (4/5) + cos-1 (12/13) = cos-1 (33/65)
By taking L.H.S
cos-1 (4/5) + cos-1 (12/13)
By Using [cos-1 x + cos-1 y = cos-1(xy - √1 - y² √1 - x²)]
= cos-1(4/5 x 12/13 - √1 - (12/13)² √1 - (4/5)²)
= cos-1(48/65 - √1 - 144/169 √1 - 16/25)
= cos-1(48/65 - √25/169 √9/25)
= cos-1[48/65 - (5/13 x 3/5)]
= cos-1[48/65 - 3/13]
= cos-1[(48-15)/65]
= cos-1(33/65)
= R.H.S
```

Question-6 :-  Prove that : cos-1 (12/13) + sin-1 (3/5) = sin-1 (56/65).

Solution :-
```   cos-1 (12/13) + sin-1 (3/5) = sin-1 (56/65)
By taking L.H.S
cos-1 (12/13) + sin-1 (3/5)
Let cos-1 (12/13) = θ
cos θ = 12/13
sin θ = √1 - (12/13)²
sin θ = √1 - 144/169
sin θ = √25/169
sin θ = 5/13
Now, θ = sin-1 (5/13)
sin-1 (5/13) + sin-1 (3/5)
By using [sin-1 x + sin-1 y = sin-1(x√1 - y² + y√1 - x²)]
= sin-1 (5/13 √1 - (3/5)² + 3/5 √1 - (5/13)²)
= sin-1 (5/13 √1 - 9/25 + 3/5 √1 - 25/169)
= sin-1 (5/13 √16/25 + 3/5 √144/169)
= sin-1 (5/13 x 4/5 + 3/5 x 12/13)
= sin-1 (4/13 + 36/65)
= sin-1 [(20 + 36)/65]
= sin-1 (56/65)
= R.H.S
```

Question-7 :-  Prove that: tan-1 (63/16) = sin-1 (5/13) + cos-1 (3/5).

Solution :-
```   tan-1 (63/16) = sin-1 (5/13) + cos-1 (3/5)
R.H.S
sin-1 (5/13) + cos-1 (3/5)
Firstly,
Let sin-1 5/13 = θ₁
sin θ₁ = 5/13
cos θ₁ = √1 - sin²θ₁
cos θ₁ = √1 - (5/13)²
cos θ₁ = √1 - 25/169
cos θ₁ = √144/169
cos θ₁ = 12/13
Now tan θ₁ = sin θ₁/cos θ₁
tan θ₁ = 5/13 x 13/12 = 5/12
θ₁ = tan-1 5/12

Secondly,
Let cos-1 (3/5) = θ₂
sin θ₂ = 3/5
sin θ₂ = √1 - cos²θ₂
sin θ₂ = √1 - (3/5)²
sin θ₂ = √1 - 9/25
sin θ₂ = √16/25
sin θ₂ = 4/5
Now tan θ₂ = sin θ₂/cos θ₂
tan θ₂ = 4/5 x 5/3 = 4/3
θ₂ = tan-1 4/3

Now, θ₁ + θ₂, We have, tan (θ₁ + θ₂) = [(tan θ₁ + tan θ₂) ÷ (1 - tan θ₁ tan θ₂)]
= tan-1 (5/12) + tan-1 (4/3)
= tan-1[(5/12 + 4/3) ÷ (1 - 5/12 x 4/3)]
= tan-1[(5 + 16)/12 ÷ (1 - 5/9)]
= tan-1[21/12 ÷ 4/9)]
= tan-1[7/4 x 9/4]
= tan-1(63/16)
= L.H.S
```

Question-8 :- Prove that: Prove that: tan-1 (1/5) + tan-1 (1/7) + tan-1 (1/3) + tan-1 (1/8) = π/4.

Solution :-
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```

Question-9 :-  Prove that : tan-1 √x = 1/2 . cos-1 (1 - x)/(1 + x), x ∈ [0,1].

Solution :-
```   tan-1 √x = 1/2 . cos-1 (1 - x)/(1 + x)
2tan-1 √x = cos-1 (1 - x)/(1 + x)
By Taking L.H.S
2tan-1 √x
We using Property, 2tan-1 x = cos-1 (1 - x²)/ (1 + x²)
= cos-1 [1 - (√x)²]/ [1 + (√x)²]
= cos-1 (1 - x)/(1 + x)
= R.H.S
```

Question-10 :-  Prove that:

Solution :-
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Question-11 :- Prove that:

Solution :-
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9

Question-12 :-  Prove that : 9π/8 - 9/4 sin-1 1/3 = 9/4 sin-1(2√2)/3.

Solution :-
```  9π/8 - 9/4 sin-1 1/3 = 9/4 sin-1(2√2)/3
Taking out common value 9/4 of both sides then,
9/4 x (π/2 - sin-1 1/3) = 9/4 sin-1(2√2)/3
π/2 - sin-1 1/3 = sin-1(2√2)/3
cos-1 1/3 = sin-1 (2√2)/3
By taking of L.H.S
let cos-1 1/3 = θ
cos θ = 1/3
sin θ = √1 - cos²θ
sin θ = √1 - (1/3)²
sin θ = √1 - 1/9
sin θ = √8/9
sin θ = (2√2)/3
θ = sin-1(2√2)/3 = R.H.S
```

Question-13 :- Solve: 2tan-1(cos x) = tan-1(2 cosec x)

Solution :-
```  2tan-1(cos x) = tan-1(2 cosec x)
tan-1[(2 cos x) ÷ (1 - cos²x)] = tan-1(2 cosec x)
(2 cos x) ÷ (1 - cos²x) = 2 cosec x
2 cos x ÷ sin²x = 2 ÷ sin x
cos x = sin x
tan x = 1
tan x = tan π/4
x = π/4
```

Question-14 :- Solve: tan-1[(1 - x) ÷ (1 + x)] = 1/2 tan-1x, (x > 0).

Solution :-
```  tan-1[(1 - x) ÷ (1 + x)] = 1/2 tan-1x
tan-1 1 - tan-1 x = 1/2 tan-1x
tan-1 (tan π/4) = 1/2 tan-1x + tan-1 x
π/4 = 3/2 tan-1x
tan-1x = π/6
x = tan (π/6)
x = 1/√3
```

Question-15 :-  sin(tan-1x), |x|<1 is equal to

Solution :-
```  sin(tan-1x)
Let tan-1x = θ
x = tan θ
sin θ = x/(√1 + x²)
θ = sin-1[x/(√1 + x²)]
tan-1x = sin-1[x/(√1 + x²)]
sin(tan-1x) = sin(sin-1[x/(√1 + x²)])
sin(tan-1x) = x/(√1 + x²)
Option is D.
```

Question-16 :-  sin-1(1 - x) - 2 sin-1x = π/2, then x is equal to

Solution :-
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Question-17 :-

Solution :-
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