NCERT Solutions for Class 12 Science Math Chapter 2 Inverse Trigonometric Functions are provided here with simple step-by-step explanations. These solutions for Inverse Trigonometric Functions are extremely popular among class 12 Science students for Math Inverse Trigonometric Functions Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the NCERT Book of class 12 Science Math Chapter 2 are provided here for you for free. You will also love the ad-free experience on Meritnationâ€™s NCERT Solutions. All NCERT Solutions for class 12 Science Math are prepared by experts and are 100% accurate.

#### Page No 41:

#### 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

#### Page No 41:

#### 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.

#### Page No 41:

#### 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

#### Page No 41:

#### 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

#### Page No 41:

#### 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

#### Page No 41:

#### 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

#### Page No 42:

#### 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

#### Page No 42:

#### 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

#### Page No 42:

#### 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

#### Page No 42:

#### 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

#### Page No 42:

#### Question 11:

Find the value of

#### Answer:

#### Page No 42:

#### Question 12:

Find the value of

#### Answer:

#### Page No 42:

#### Question 13:

Find the value of if
sin^{−1}*x *= *y*,
then

**(****A)** **(B)**

**(****C)** **(D) **

#### Answer:

It
is given that sin^{−1}*x *= *y*.

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

Therefore,.

#### Page No 42:

#### Question 14:

Find
the value of **
**is
equal to

**(****A)** π (**B)** (**C)** (**D) **

#### Answer:

#### Page No 47:

#### Question 1:

Prove

#### Answer:

To prove:

Let *x* = sin*θ*.
Then,

We have,

R.H.S. =

=
3*θ*

= L.H.S.

#### Page No 47:

#### Question 2:

Prove

#### Answer:

To prove:

Let *x* = cos*θ*.
Then, cos^{−1}*x* =*θ*.

We have,

#### Page No 47:

#### Question 3:

Prove

#### Answer:

To prove:

#### Page No 47:

#### Question 4:

Prove

#### Answer:

To prove:

#### Page No 47:

#### Question 5:

Write the function in the simplest form:

#### Answer:

#### Page No 47:

#### Question 6:

Write the function in the simplest form:

#### Answer:

Put *x* = cosec *θ*
⇒
*θ* = cosec^{−1}*x*

#### Page No 47:

#### Question 7:

Write the function in the simplest form:

#### Answer:

#### Page No 47:

#### Question 8:

Write the function in the simplest form:

#### Answer:

${\mathrm{tan}}^{-1}\left(\frac{\mathrm{cos}x-\mathrm{sin}x}{\mathrm{cos}x+\mathrm{sin}x}\right)\phantom{\rule{0ex}{0ex}}={\mathrm{tan}}^{-1}\left(\frac{1-{\displaystyle \frac{\mathrm{sin}x}{\mathrm{cos}x}}}{1+{\displaystyle \frac{\mathrm{sin}x}{\mathrm{cos}x}}}\right)\phantom{\rule{0ex}{0ex}}={\mathrm{tan}}^{-1}\left(\frac{1-\mathrm{tan}x}{1+\mathrm{tan}x}\right)\phantom{\rule{0ex}{0ex}}={\mathrm{tan}}^{-1}\left(1\right)-{\mathrm{tan}}^{-1}\left(\mathrm{tan}x\right)\left({\mathrm{tan}}^{-1}\frac{x-y}{1+xy}={\mathrm{tan}}^{-1}x-{\mathrm{tan}}^{-1}y\right)\phantom{\rule{0ex}{0ex}}=\frac{\mathrm{\pi}}{4}-x$

#### Page No 48:

#### Question 9:

Write the function in the simplest form:

#### Answer:

#### Page No 48:

#### Question 10:

Write the function in the simplest form:

#### Answer:

#### Page No 48:

#### Question 11:

Find the value of

#### Answer:

Let. Then,

#### Page No 48:

#### Question 12:

Find the value of

#### Answer:

#### Page No 48:

#### 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*.

#### Page No 48:

#### Question 14:

If,
then find the value of *x*.

#### Answer:

On squaring both sides, we get:

Hence, the value of *x*
is

#### Page No 48:

#### Question 15:

If,
then find the value of *x*.

#### Answer:

Hence, the value of *x*
is

#### Page No 48:

#### Question 16:

Find the values of

#### Answer:

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

Here,

Now, can be written as:

#### Page No 48:

#### Question 17:

Find the values of

#### Answer:

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

Here,

Now, can be written as:

#### Page No 48:

#### Question 18:

Find the values of

#### Answer:

Let. Then,

#### Page No 48:

#### 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 ^{−1}*x*.

Here,

Now, can be written as:

${\mathrm{cos}}^{-1}\left(\mathrm{cos}\frac{7\mathrm{\pi}}{6}\right)={\mathrm{cos}}^{-1}\left[\mathrm{cos}\left(\mathrm{\pi}+\frac{\mathrm{\pi}}{6}\right)\right]\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}{\mathrm{cos}}^{-1}\left(\mathrm{cos}\frac{7\mathrm{\pi}}{6}\right)={\mathrm{cos}}^{-1}\left[-\mathrm{cos}\frac{\mathrm{\pi}}{6}\right]\left[\mathrm{as},\mathrm{cos}\left(\mathrm{\pi}+\mathrm{\theta}\right)=-\mathrm{cos}\mathrm{\theta}\right]\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}{\mathrm{cos}}^{-1}\left(\mathrm{cos}\frac{7\mathrm{\pi}}{6}\right)={\mathrm{cos}}^{-1}\left[-\mathrm{cos}\left(\mathrm{\pi}-\frac{5\mathrm{\pi}}{6}\right)\right]\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}{\mathrm{cos}}^{-1}\left(\mathrm{cos}\frac{7\mathrm{\pi}}{6}\right)={\mathrm{cos}}^{-1}\left[-\left\{-\mathrm{cos}\left(\frac{5\mathrm{\pi}}{6}\right)\right\}\right]\left[\mathrm{as},\mathrm{cos}\left(\mathrm{\pi}-\mathrm{\theta}\right)=-\mathrm{cos}\mathrm{\theta}\right]\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

The correct answer is B.

#### Page No 48:

#### 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.

#### Page No 48:

#### 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.

#### Page No 51:

#### Question 1:

Find the value of

#### Answer:

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

Here,

Now, can be written as:

#### Page No 51:

#### Question 2:

Find the value of

#### Answer:

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

Here,

Now, can be written as:

#### Page No 51:

#### Question 3:

Prove

#### Answer:

Now, we have:

#### Page No 51:

#### Question 4:

Prove

#### Answer:

Now, we have:

#### Page No 51:

#### Question 5:

Prove

#### Answer:

Now, we will prove that:

#### Page No 51:

#### Question 6:

Prove

#### Answer:

Now, we have:

#### Page No 51:

#### Question 7:

Prove

#### Answer:

Using (1) and (2), we have

#### Page No 51:

#### Question 8:

Prove

#### Answer:

#### Page No 52:

#### Question 9:

Prove

#### Answer:

#### Page No 52:

#### Question 10:

Prove

#### Answer:

#### Page No 52:

#### Question 11:

Prove
[**Hint: **put*x* =
cos 2*θ*]

#### Answer:

#### Page No 52:

#### Question 12:

Prove

#### Answer:

#### Page No 52:

#### Question 13:

Solve

#### Answer:

#### Page No 52:

#### Question 14:

Solve

#### Answer:

#### Page No 52:

#### Question 15:

Solveis equal to

**(A)
**
(**B) **
(**C)
**
(**D)
**

#### Answer:

Let
tan^{−1}*x* = *y*.
Then,

The correct answer is D.

#### Page No 52:

#### Question 16:

Solve**,
**then *x*
is equal to

**(****A)
**
(**B)
**
(**C)**
0 (**D)
**

#### Answer:

Therefore, from equation (1), we have

Put
*x* = sin
*y*. Then,
we have:

But, when, it can be observed that:

is not the solution of the given equation.

Thus,
*x* = 0.

Hence,
the correct answer is **C**.

#### Page No 52:

#### Question 17:

Solveis equal to

**(A)**
**(B).**
**(C)**
**(D)
**

#### Answer:

Hence, the correct answer is **C**.

View NCERT Solutions for all chapters of Class 12