2 – INVERSE TRIGONOMETRIC FUNCTIONS


Here we are going through the ncert solutions for class 12 maths chapter 2 – inverse trigonometric functions so before going through the ncert solutions make sure to go through the textbook that helps you to understand the solutions more easily

EXERCISE 2.1

 

Question 1:

Find the principal value of 

ANSWER:

Let sin-1  Then sin y = 

We know that the range of the principal value branch of sin−1 is

 and sin

Therefore, the principal value of 

 

Question 2:

Find the​​ principal value of 

ANSWER:

We know that the range of the principal value branch of cos−1 is

.

Therefore, the principal value of.

 

Question 3:

Find the principal value of cosec−1 (2)

ANSWER:

Let cosec−1 (2) = y. Then, 

We know that the range of​​ the principal value branch of cosec−1 is 

Therefore, the principal value of 

 

Question 4:

Find the principal value of 

ANSWER:

We know that the range of the principal value branch of tan−1 is 

Therefore, the principal value of 

 

Question 5:

Find​​ the principal value of 

ANSWER:

We know that the range of the principal value branch of cos−1 is

Therefore, the principal value of 

 

Question 6:

Find the principal value of tan−1 (−1)

ANSWER:

Let tan−1 (−1) = y. Then, 

We know that the range of​​ the principal value branch of tan−1 is

Therefore, the principal value of 



 

Question 7:

Find the principal value of 

ANSWER:

We know that the range of the principal value branch of sec−1 is

Therefore, the principal value of 

 

Question 8:

Find the principal value of 

ANSWER:

We know that the range of the principal value branch of cot−1 is (0,π) and

Therefore, the principal value of 

 

Question 9:

Find the principal value of 

ANSWER:

We know that the range of the principal value​​ branch of cos−1 is [0,π] and

.

Therefore, the principal value of 

 

Question 10:

Find the principal value of 

ANSWER:

We know that the range of the principal value branch of cosec−1 is 

Therefore, the principal value of 

 

Question 11:

Find the​​ value of 

ANSWER:

 

Question 12:

Find the value of 

ANSWER:

 

Question 13:

Find the value of if sin−1 y, then

(A)  (B) 

(C)  (D) 

ANSWER:

It is given that sin−1 y.

We know that the range of the principal value branch of sin−1 is 

Therefore,.

 

Question 14:

Find the value of is equal to

(A) π (B)  (C)  (D) 

ANSWER:


​​  ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​​​ EXERCISE 2.2

 

Question 1:

Prove 

ANSWER:

To prove: 

Let x = sinθ. Then, 

We have,

R.H.S. =

= 3θ

=​​ L.H.S.

 

Question 2:

Prove 

ANSWER:

To prove:

Let x = cosθ. Then, cos−1 x =θ.

We have,

 

Question 3:

Prove 

ANSWER:

To prove:

 

Question 4:

Prove 

ANSWER:

To prove: 

 

Question 5:

Write the function in the simplest form:

ANSWER:

 

Question 6:

Write the function in the simplest form:

ANSWER:

Put x = cosec θ  θ = cosec−1 x

 

Question 7:

Write the function in the simplest form:

ANSWER:

 

Question 8:

Write the function in the simplest form:

ANSWER:

tan−1(cosx−sinxcosx+sinx)=tan−1(1−sinxcosx1+sinxcosx)=tan−1(1−tanx1+tanx)=tan−1(1)−tan−1(tanx)        (tan−1xy1+xy=tan−1x−tan−1y)=π4xtan-1cosx-sinxcosx+sinx=tan-11-sinxcosx1+sinxcosx=tan-11-tanx1+tanx=tan-11-tan-1tanx        tan-1x-y1+xy=tan-1x-tan-1y=π4-x



 

Question 9:

Write the function in the simplest form:

ANSWER:

 

Question 10:

Write the function in the simplest form:

ANSWER:

 

Question 11:

Find the value of 

ANSWER:

Let. Then,

 

Question 12:

Find the value of 

ANSWER:

 

Question 13:

Find the value of 

ANSWER:

Let x = tan θ. Then, θ = tan−1 x.

Let y = tan Φ. Then, Φ = tan−1 y.

 

Question 14:

If, then​​ find the value of x.

ANSWER:

On squaring both sides, we get:

Hence, the value of x is

 

Question 15:

If, then find the value of x.

ANSWER:

Hence, the value of x is 

 

Question 16:

Find the values of 

ANSWER:

We know that sin−1 (sin x) = x if,​​ which is the principal value branch of sin−1x.

Here,

Now, can be written as:

 

Question 17:

Find the values of 

ANSWER:

We know that tan−1 (tan x) = x if, which is the principal value branch of tan−1x.

Here,

Now, can be written as:

 

Question 18:

Find the values of 

ANSWER:

Let. Then,

 

Question 19:

Find the values of is equal to

(A)  (B)  (C)  (D) 

ANSWER:

We know that cos−1 (cos x) = x if, which is the principal value branch of cos −1x.

Here,

Now, can be written as:


cos−1(cos7π6) = cos−1[cos(π+π6)]cos−1(cos7π6) = cos−1[− cosπ6]             [as, cos(π+θ) = − cos θ]cos−1(cos7π6)  = cos−1[− cos(π5π6)]cos−1(cos7π6) = cos−1[{− cos (5π6)}]   [as, cos(πθ) = − cos θ]cos-1cos7π6 = cos-1cosπ+π6cos-1cos7π6 = cos-1- cosπ6             as, cosπ+θ = - cos θcos-1cos7π6  = cos-1- cosπ-5π6cos-1cos7π6 = cos-1-- cos 5π6   as, cosπ-θ = - cos θ

The correct answer is B.

 

Question 20:

Find the values of is equal to

(A)  (B)  (C)  (D) 1

ANSWER:

Let. Then, 

We know that the range of the principal value branch of.

The correct answer is D.

 

Question 21:

Find the values of is equal to

(A) π (B)  (C) 0 (D) 

ANSWER:

Let. Then,

We know that the range of the​​ principal value branch of

Let.

The range of the principal value branch of

The correct answer is B.



 ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​​​ Miscellaneous Exercise on Chapter 2

 

Question 1:

Find the value of 

ANSWER:

We know that cos−1 (cos x) = x if,​​ which is the principal value branch of cos −1x.

Here,

Now, can be written as:

 

Question 2:

Find the value of 

ANSWER:

We know that tan−1 (tan x) = x if, which is the principal value branch of tan −1x.

Here,

Now, can be written as:

 

Question 3:

Prove 

ANSWER:

Now, we have:

 

Question 4:

Prove 

ANSWER:

Now, we have:

 

Question 5:

Prove 

ANSWER:

Now, we will prove that:

 

Question 6:

Prove 

ANSWER:

Now, we have:

 

Question 7:

Prove 

ANSWER:

Using (1) and (2), we​​ have