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# Class 12 NCERT Solutions- Mathematics Part I – Chapter 2 Inverse Trigonometric Functions – Exercise 2.2 | Set 1

### Question 1. 3sin-1x = sin-1(3x – 4x3), x∈[-1/2, 1/2]

Solution:

Let us take x = sinθ, so θ = sin-1x

Substitute the value of x in the equation present on R.H.S.

The equation becomes sin-1(3sinθ – 3sin3θ)

We know, sin3θ = 3sinθ – 4sin3θ

So , sin-1(3sinθ – 3sin3θ) = sin-1(sin3θ)

By the property of inverse trigonometry we know, sin(sin-1(θ)) = θ

So, sin-1(sin3θ) = 3θ

And we know θ = sin-1x

So, 3θ = 3sin-1x = L.H.S

### Question 2. 3cos-1x = cos-1(4x3 – 3x), x∈[-1/2, 1]

Solution:

Let us take x = cosθ, so θ = cos-1x

Substitute value of x in the equation present on R.H.S.

The equation becomes cos-1(4cos3θ – 3cosθ)

We know, cos3θ = 4cos3θ – 3cosθ

So, cos-1(4cos3θ – 3cosθ) = cos-1(cos3θ)

By the property of inverse trigonometry we know, cos(cos-1(θ)) = θ

So, cos-1(cos3θ) = 3θ

And we know θ = cos-1x

So, 3θ = 3cos-1x = 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-1x in terms of  tan-1x

We know that 2tan-1x =

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

### Question 5.

Solution:

Let us assume that x = tanθ, so θ = tan-1

Substitute the value of x in question.

So equation becomes

We know that, 1 + tan2θ = sec2θ

Replacing 1 + tan2θ with sec2θ in the equation

So equation becomes,

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

Replacing value of tanθ and secθ in

We know, 1 – cosθ = 2sin2θ/2​ and sinθ = 2sinθ/2cosθ/2

So the equations after replacing above value becomes

We know

= θ/2           [tan-1(tanθ) = θ]

= 1/2 tan-1x            [θ = tan-1x]

### Question 6.  , |x| > 1

Solution:

Let us assume that x = cosecθ, so θ = cosec -1

Substitute the value of x in question with

We know that, 1 + cot2θ = cosec2θ, so cosec2θ = 1 – cot2θ

= tan-1(tanθ)          [1/cotθ = tanθ]

= θ           [tan-1(tanθ) = θ]

= cosec−1x          [θ = cosec−1x]

= π/2 ​- sec−1x         [cosec−1x + sec−1x = π/2​]

### Question 7.

Solution:

We know, 1 – cosx = 2sin2x/2 and 1 + cosx = 2cos2x/2

Substituting 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-1x

So we can say that

= π/4​ – tan−1x          [tan−11 = π/4​]

### Question 9.

Solution:

Let us assume that x = asinθ, so θ = sin -1x/a

Substitute the value of x in question.

Taking a2 common from denominator

We know that, sin2θ + cos2θ = 1, so 1 – sin2θ = cos2θ

= tan-1(tanθ)          [sinθ/cosθ = tanθ]

= θ

= sin-1x/a

### Question 10.

Solution:

Let us assume that x = atanθ, so θ = tan -1x/a

Substitute the value of x in question

Taking a3common from numerator and denominator

We know

So,

= 3θ           [ tan-1(tanθ) = θ]

= 3tan -1x/a

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

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