﻿ Class 12 NCERT Math Solution
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TOPICS
Exercise - 2.2

Question-1 :-  Prove that 3sin-1 x = sin-1 (3x - 4x³), x ∈ [-1/2, 1/2].

Solution :-
```   Let x = sin θ and θ = sin-1 x.
R.H.S
sin-1 (3x - 4x³)
= sin-1 (3sin θ - 4sin³ θ)
= sin-1 (sin 3θ)
= 3θ = 3sin-1 x
```

Question-2 :-  Prove that 3cos-1 x = cos-1 (4x³ - 3x), x ∈ [1/2, 1].

Solution :-
```   Let x = cos θ and θ = cos-1 x.
R.H.S
cos-1 (4x³ - 3x)
= cos-1 (4cos³ θ - 3cos θ)
= cos-1 (cos 3θ)
= 3θ = 3cos-1 x
```

Question-3 :-  Prove that .

Solution :-
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Question-4 :-  Prove that .

Solution :-
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Question-5 :-  Write the function in simplest form : .

Solution :-
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Question-6 :-  Write the function in simplest form : .

Solution :-
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Question-7 :-  Write the function in simplest form : .

Solution :-
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Question-8 :-  Write the function in simplest form : .

Solution :-
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Question-9 :-  Write the function in simplest form : .

Solution :-
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Question-10 :-  Write the function in simplest form : .

Solution :-
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Question-11 :-  Find the value of.

Solution :-
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Question-12 :-  Find the value of cot (tan-1a + cot-1a).

Solution :-
```   cot(tan-1a + cot-1a)
= cot(π/2)   ∴ [tan-1a + cot-1a = π/2]
= 0
```

Question-13 :-  Find the value of.

Solution :-
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Question-14 :-  Find the value of x, if sin (sin-1 1/5 + cos-1 x) = 1.

Solution :-
```    sin (sin-1 1/5 + cos-1 x) = 1
sin (sin-1 1/5 + cos-1 x) = sin π/2
sin-1 1/5 + cos-1 x = sin-1 (sin π/2)
sin-1 1/5 + cos-1 x  = π/2    ∴ [sin-1x + cos-1x = π/2]
sin-1 1/5 = π/2 - cos-1 x
sin-1 1/5 = sin-1 x
1/5 = x or x = 1/5
```

Question-15 :-  Find the value of.

Solution :-
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Question-16 :-  Find the value of sin-1(sin 2π/3).

Solution :-
```   sin-1(sin 2π/3)
Range of sin-1 is [-π/2, π/2], so 2π/3 is not in Range. Then we break 2π/3.
= sin-1[sin(π - π/3)]
= sin-1[sin π/3]
= π/3
```

Question-17 :-  Find the value of tan-1(tan 3π/4).

Solution :-
```   tan-1(tan 3π/4)
Range of tan-1 is [-π/2, π/2], so 3π/4 is not in Range. Then we break 3π/4.
= tan-1[tan(π - π/4)]
= tan-1[-tan π/4]
= tan-1[tan (-π/4)]
= -π/4
```

Question-18 :-  Find the value of tan(sin-1 3/5 + cot-1 3/2).

Solution :-
```   tan(sin-1 3/5 + cot-1 3/2)

Firstly,
Let sin-1 3/5 = θ₁
sin θ₁ = 3/5
cos θ₁ = √1 - sin²θ₁
cos θ₁ = √1 - (3/5)²
cos θ₁ = √1 - 9/25
cos θ₁ = √16/25
cos θ₁ = 4/5
Then, tan θ₁ = sin θ₁/cos θ₁ = 3/5 x 5/4 = 3/4
tan θ₁ = 3/4
θ₁ = tan-1 3/4

Secondly,
Let cot-1 3/2 = θ₂
cot θ₂ = 3/2
tan θ₂ = 1/cot θ₂
tan θ₂ = 2/3
θ₂ = tan-1 2/3

Now, Accordin to question
tan(θ₁ + θ₂)
= tan(tan-1 3/4 + tan-1 2/3)
= tan[tan-1(3/4 + 2/3) ÷ (1 - 3/4 x 2/3)]
= tan[tan-1(9 + 8)/12 ÷ (1 - 1/2)]
= tan[tan-1 (17/12 ÷ 1/2)]
= tan[tan-1 (17/12 x 2/1)]
= tan[tan-1 (17/6)]
= 17/6
```

Question-19 :-  cos-1(cos 7π/6) is equal to
(A) 7π/6   (B) 5π/6  (C) π/3  (D) π/6

Solution :-
```   cos-1(cos 7π/6)
Range of cos-1 is [0, π], so 7π/6 is not in Range. Then we break 7π/6.
= cos-1[cos(π + π/6)]
= cos-1[-cos π/6]
= cos-1[cos (-π/6)]  {Range = [0, π]}
= cos-1[cos (π - π/6)]
= cos-1[cos (5π/6)]
= 5π/6
Option is (B)
```

Question-20 :-  sin [π/3 + sin-1(-1/2)] is equal to
(A) 1/2   (B) 1/3  (C) 1/4  (D) 1

Solution :-
```   sin [π/3 + sin-1(-1/2)]
= sin [π/3 + sin-1(-sin π/6)]
= sin [π/3 + sin-1(sin (-π/6))]
= sin [π/3 - π/6)]
= sin [(2π - π)/6]
= sin [π/6]
= 1/2
```

Question-21 :-  tan-1(√3) - cot-1(-√3) is equal to
(A) π   (B) -π/2  (C) 0  (D) 2√3

Solution :-
```   tan-1(√3) - cot-1(-√3)
Let tan-1(√3) = θ₁
tan θ₁ = √3      {Range = [-π/2, π/2]}
tan θ₁ = tan π/3
θ₁ = π/3

Let cot-1(-√3) = θ₂
cot θ₂ = -√3
cot θ₂ = -cot π/6
cot θ₂ = cot (-π/6)  {Range = [0, π]}
cot θ₂ = cot (π - π/6)
cot θ₂ = cot (5π/6)
θ₂ = 5π/6

Now, θ₁ - θ₂
= π/3 - 5π/6
= (2π - 5π)/6
= -3π/6
= -π/2
Option (B) is correct.
```
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