NCERT Solutions Class 12 maths Chapter-2 (Inverse Trigonometric Functions)Exercise 2.2

NCERT Solutions Class 12 Maths from class 12th Students will get the answers of Chapter-2 (Inverse Trigonometric Functions)Exercise 2.2 This chapter will help you to learn the basics and you should expect at least one question in your exam from this chapter.
We have given the answers of all the questions of NCERT Board Mathematics Textbook in very easy language, which will be very easy for the students to understand and remember so that you can pass with good marks in your examination.

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

Exercise 2.2  

Set 2

Find the values of each of the following: 

Question 11. tan⁻¹[2cos(2sin⁻¹1/2​)]

Solution:

Let us assume that sin⁻¹1/2 = x

So, sinx = 1/2

Therefore, x = π​/6 = sin⁻¹1/2

Therefore, tan⁻¹[2cos(2sin⁻¹1/2​)] =  tan⁻¹[2cos(2 * π​/6)]

= tan⁻¹[2cos(π​/3)]

Also, cos(π/3​) = 1/2​

Therefore, tan⁻¹[2cos(π​/3)] = tan⁻¹[(2 * 1/2)]

= tan⁻¹[1] = π​/4 

Question 12. cot(tan⁻¹a + cot⁻¹a) 

Solution:

We know, tan⁻¹x + cot⁻¹x = π​/2

Therefore, cot(tan⁻¹a + cot⁻¹a) = cot(π​/2) =0

Question 13.  

Solution:

We know, 2tan-1x =  and 2tan-1y =  

= tan(1/2)​[2(tan−1x + tan−1y)]

= tan[tan−1x + tan−1y]

Also, tan−1x + tan−1y = 

Therefore, tan[tan−1x + tan−1y] = 

= (x + y)/(1 – xy)

Question 14. If sin(sin⁻¹1/5​+ cos⁻¹x) = 1 then find the value of x

Solution:

sin⁻¹1/5​ + cos⁻¹x = sin⁻¹1

We know, sin⁻¹1 = π/2

Therefore, sin⁻¹1/5​ + cos⁻¹x = π/2

sin⁻¹1/5​ = π/2 – cos⁻¹x

Since, sin⁻¹x​ + cos⁻¹x = π/2

Therefore, π/2 – cos⁻¹x = sin⁻¹x

sin⁻¹1/5​ = sin⁻¹x

So, x = 1/5

Question 15. If  , then find the value of x

Solution:

We know, tan−1x + tan−1y = 

2x2 – 4 = -3

2x2 – 4 + 3 = 0

2x2 – 1 = 0

x2 = 1/2

x = 1/√2, -1/√2

Find the values of each of the expressions in Exercises 16 to 18.

Question 16. sin ^{− 1}(sin2π/3​)  

Solution:

We know that sin−1(sinθ) = θ when θ ∈ [-π/2, π/2], but 

So, sin − 1(sin2π/3​) can be written as 

 sin − 1(sinπ/3​)  here 

Therefore, sin − 1(sinπ/3​) = π/3

Question 17. tan⁻¹(tan3π/4​)

Solution:

We know that tan−1(tanθ) = θ when  but 

So, tan−1(tan3π/4​) can be written as tan−1(-tan(-3π/4)​)

= tan−1[-tan(π – π/4​)]

= tan−1[-tan(π/4​)]

= –tan−1[tan(π/4​)]

= – π/4 where 

Question 18.

Solution:

Let us assume  = x , so sinx = 3/5 

We know, 

cosx = 4/5

We know, 

So, 

tanx = 3/4

Also, 

Hence, 

tan-1x + tan-1y = 

So, 

= 17/6

Question 19.  cos⁻¹(cos7π/6​) is equal to

(i) 7π/6    (ii) 5π/6    (iii)π/3    (iv)π/6

Solution:

 We know that cos⁻¹(cosθ) = θ, θ ∈ [0, π]

cos⁻¹(cosθ) = θ, θ ∈ [0, π]

Here, 7π/6 > π 

So, cos⁻¹(cos7π/6​) can be written as cos⁻¹(cos(-7π/6)​)

= cos⁻¹[cos(2π – 7π/6​)]      [cos(2π + θ) = θ]

= cos⁻¹[cos(5π/6​)]       where 5π/6 ∈  [0, π]

  Therefore, cos⁻¹[cos(5π/6​)] = 5π/6 

Question 20. 
(i) 1/2    (ii) 1/3   (iii) 1/4    (iv) 1

Solution:

Let us assume sin^-1(-1/2)= x, so sinx = -1/2 

Therefore, x = -π/6​

Therefore, sin[π/3​ – (-π/6​)]

= sin[π/3​ + (π/6​)]

= sin[3π/6]

= sin[π/2]

= 1

Question 21.  is equal to
(i) π    (ii) -π/2    (iii)0    (iv)2√3

Solution:

We know, cot(−x) = −cotx

Therefore, tan^-13 – cot^-1(-3) = tan^-13 – [-cot^-1(3)]

= tan^-13 + cot^-13

Since, tan^-1x + cot^-1x = π/2

Tan^-13 + cot^-13 = -π/2

Chapter-2 (Inverse Trigonometric Functions)

• Exercise 2.1