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

Prove the following

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

Write the following functions in simplest forms: 

Question 5.  

Solution:

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

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 ^-1x 

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