### Prove the following

### Question 1. 3sin^{-1}x = sin^{-1}(3x – 4x^{3}), x∈[-1/2, 1/2]

**Solution:**

Let us take x = sinθ, so θ = sin

^{-1}xSubstitute the value of x in the equation present on R.H.S.

The equation becomes sin

^{-1}(3sinθ – 3sin^{3}θ)We know, sin3θ = 3sinθ – 4sin

^{3}θSo , sin

^{-1}(3sinθ – 3sin^{3}θ) = sin^{-1}(sin3θ)By the property of inverse trigonometry we know, sin(sin

^{-1}(θ)) = θSo, sin

^{-1}(sin3θ) = 3θAnd we know θ = sin

^{-1}xSo, 3θ = 3sin

^{-1}x = L.H.S

### Question 2. 3cos^{-1}x = cos^{-1}(4x^{3} – 3x), x∈[-1/2, 1]

**Solution:**

Let us take x = cosθ, so θ = cos

^{-1}xSubstitute value of x in the equation present on R.H.S.

The equation becomes cos

^{-1}(4cos^{3}θ – 3cosθ)We know, cos3θ = 4cos

^{3}θ – 3cosθSo, cos

^{-1}(4cos^{3}θ – 3cosθ) = cos^{-1}(cos3θ)By the property of inverse trigonometry we know, cos(cos

^{-1}(θ)) = θSo, cos

^{-1}(cos3θ) = 3θAnd we know θ = cos

^{-1}xSo, 3θ = 3cos

^{-1}x = L.H.S

### Question 3.

**Solution:**

We know,

Now put x = 2/11 and y = 7/24

So,

= R.H.S

### Question 4.

**Solution:**

We have to first write 2tan

^{-1}x in terms of tan^{-1}xWe know that 2tan

^{-1}x =Put x = 1/2 in the above formula

So,

Now we can replace with

So equation in L.H.S become

We know ,

So,

= R.H.S

### Write the following functions in simplest forms:

### Question 5.

**Solution:**

Let us assume that x = tanθ, so θ = tan

^{-1}xSubstitute the value of x in question.

So equation becomes

We know that, 1 + tan

^{2}θ = sec^{2}θReplacing 1 + tan

^{2}θ with sec^{2}θ in the equationSo equation becomes,

We know, tanθ = sinθ/cosθ and sec = 1/cosθ

Replacing value of tanθ and secθ in

We know, 1 – cosθ = 2sin

^{2}θ/2 and sinθ = 2sinθ/2cosθ/2So the equations after replacing above value becomes

We know

= θ/2 [tan

^{-1}(tanθ) = θ]= 1/2 tan

^{-1}x [θ = tan^{-1}x]

### Question 6. , |x| > 1

**Solution:**

Let us assume that x = cosecθ, so θ = cosec

^{-1}xSubstitute the value of x in question with

We know that, 1 + cot

^{2}θ = cosec^{2}θ, so cosec^{2}θ = 1 – cot^{2}θ= tan

^{-1}(tanθ) [1/cotθ = tanθ]= θ [tan

^{-1}(tanθ) = θ]= cosec

^{−1}x [θ = cosec^{−1}x]= π/2 - sec

^{−1}x [cosec^{−1}x + sec^{−1}x = π/2]

### Question 7.

**Solution:**

We know, 1 – cosx = 2sin

^{2}x/2 and 1 + cosx = 2cos^{2}x/2Substituting above formula in question

= tan

^{-1}(tanx/2)= x/2 [tan

^{-1}(tanθ) = θ]

### Question 8.

**Solution:**

Divide numerator and denominator by

We know,

This can also be written as – (1)

We know – (2)

On comparing equation (1) and (2) we can say that x = 1 and y = tan

^{-1}xSo we can say that

= π/4 – tan

^{−1}x [tan^{−1}1 = π/4]

### Question 9.

**Solution:**

Let us assume that x = asinθ, so θ = sin

^{-1}x/aSubstitute the value of x in question.

Taking a

^{2 }common from denominatorWe know that, sin

^{2}θ + cos^{2}θ = 1, so 1 – sin^{2}θ = cos^{2}θ= tan

^{-1}(tanθ) [sinθ/cosθ = tanθ]= θ

= sin

^{-1}x/a

### Question 10.

**Solution:**

Let us assume that x = atanθ, so θ = tan

^{-1}x/aSubstitute the value of x in question

Taking a

^{3}common from numerator and denominatorWe know

So,

= 3θ [ tan

^{-1}(tanθ) = θ]= 3tan

^{-1}x/a

### Chapter 2 Inverse Trigonometric Functions – Exercise 2.2 | Set 2

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