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

Last Updated : 01 May, 2024

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

Solution:

Let us take x = sinÎ¸, so Î¸ = sin-1x

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

The equation becomes sin-1(3sinÎ¸ – 3sin3Î¸)

We know, sin3Î¸ = 3sinÎ¸ – 4sin3Î¸

So , sin-1(3sinÎ¸ – 3sin3Î¸) = sin-1(sin3Î¸)

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

So, sin-1(sin3Î¸) = 3Î¸

And we know Î¸ = sin-1x

So, 3Î¸ = 3sin-1x = L.H.S

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

Solution:

Let us take x = cosÎ¸, so Î¸ = cos-1x

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

The equation becomes cos-1(4cos3Î¸ – 3cosÎ¸)

We know, cos3Î¸ = 4cos3Î¸ – 3cosÎ¸

So, cos-1(4cos3Î¸ – 3cosÎ¸) = cos-1(cos3Î¸)

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

So, cos-1(cos3Î¸) = 3Î¸

And we know Î¸ = cos-1x

So, 3Î¸ = 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 = tanÎ¸, so Î¸ = tan-1

Substitute the value of x in question.

So equation becomes

We know that, 1 + tan2Î¸ = sec2Î¸

Replacing 1 + tan2Î¸ with sec2Î¸ in the equation

So equation becomes,

We know, tanÎ¸ = sinÎ¸/cosÎ¸ and sec = 1/cosÎ¸

Replacing value of tanÎ¸ and secÎ¸ in

We know, 1 – cosÎ¸ = 2sin2Î¸/2â€‹ and sinÎ¸ = 2sinÎ¸/2cosÎ¸/2

So the equations after replacing above value becomes

We know

= Î¸/2           [tan-1(tanÎ¸) = Î¸]

= 1/2 tan-1x            [Î¸ = tan-1x]

### Question 6.  , |x| > 1

Solution:

Let us assume that x = cosecÎ¸, so Î¸ = cosec -1

Substitute the value of x in question with

We know that, 1 + cot2Î¸ = cosec2Î¸, so cosec2Î¸ = 1 – cot2Î¸

= tan-1(tanÎ¸)          [1/cotÎ¸ = tanÎ¸]

= Î¸           [tan-1(tanÎ¸) = Î¸]

= cosecâˆ’1x          [Î¸ = cosecâˆ’1x]

= Ï€/2 â€‹- secâˆ’1x         [cosecâˆ’1x + secâˆ’1x = Ï€/2â€‹]

### 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(tanÎ¸) = Î¸]

### 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

= Ï€/4â€‹ – tanâˆ’1x          [tanâˆ’11 = Ï€/4â€‹]

### Question 9.

Solution:

Let us assume that x = asinÎ¸, so Î¸ = sin -1x/a

Substitute the value of x in question.

Taking a2 common from denominator

We know that, sin2Î¸ + cos2Î¸ = 1, so 1 – sin2Î¸ = cos2Î¸

= tan-1(tanÎ¸)          [sinÎ¸/cosÎ¸ = tanÎ¸]

= Î¸

= sin-1x/a

### Question 10.

Solution:

Let us assume that x = atanÎ¸, so Î¸ = tan -1x/a

Substitute the value of x in question

Taking a3common from numerator and denominator

We know

So,

= 3Î¸           [ tan-1(tanÎ¸) = Î¸]

= 3tan -1x/a

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