# Exercise 2.1 Solutions

This page contains the NCERT mathematics class 12 chapter Inverse Trigonometric Functions Exercise 2.1 Solutions. You can find the numerical questions solutions for the chapter 2/Exercise 2.1 of NCERT class 12 mathematics in this page. So is the case if you are looking for NCERT class 12 Maths related topic Inverse Trigonometric Functions. This page contains Exercise 2.1 solutions. If you’re looking for summary of the chapter or other exercise solutions, they’re available at
Exercise 2.1 Solutions
1. Find the value of the following: {\sin^{-1}\left(-\dfrac{1}{2}\right)}
Let {\sin^{-1}\left(-\dfrac{1}{2}\right) = y}. Then,
{\sin y}
=
-\dfrac{1}{2}
=
{\sin\left(-\dfrac{π}{6}\right)}
We know that the range of principal value branch of sin-1 is {\left[-\dfrac{π}{2}, \dfrac{π}{2}\right]} and {\sin\left(-\dfrac{π}{6}\right) = -\dfrac{1}{2}}.
∴ The principal value of {\sin^{-1}\left(-\dfrac{1}{2}\right)} is {-\dfrac{π}{6}}
2. Find the value of the following: {\cos^{-1}\left(\dfrac{\sqrt{3}}{2}\right)}
Let {\cos^{-1}\left(\dfrac{\sqrt{3}}{2}\right) = y}. Then,
{\cos y}
=
\dfrac{\sqrt{3}}{2}
=
{\cos\left(\dfrac{π}{6}\right)}
We know that the range of principal value branch of cos-1 is {[0, π]} and {\cos\left(\dfrac{π}{6}\right) = \dfrac{\sqrt{3}}{2}}.
∴ The principal value of {\cos^{-1}\left(\dfrac{\sqrt{3}}{2}\right)} is {\dfrac{π}{6}}
3. Find the value of the following: cosec-1(2)
Let {\cosec^{-1}(2) = y}. Then,
{\cosec y}
=
2
=
{\cosec\left(\dfrac{π}{6}\right)}
We know that the range of principal value branch of cosec-1 is {\left[-\dfrac{π}{2}, \dfrac{π}{2}\right] - \{0\}} and {\cosec\left(\dfrac{π}{6}\right) = 2}.
∴ The principal value of {\cosec^{-1}(2)} is {\dfrac{π}{6}}
4. Find the value of the following: {\tan^{-1}\left(-\sqrt{3}\right)}
Let {\tan^{-1}\left(-\sqrt{3}\right) = y}. Then,
{\tan y}
=
-\sqrt{3}
=
{\tan\left(-\dfrac{π}{3}\right)}
We know that the range of principal value branch of tan-1 is {\left(-\dfrac{π}{2}, \dfrac{π}{2}\right)} and {\tan\left(-\dfrac{π}{3}\right) = -\sqrt{3}}.
∴ The principal value of {\tan^{-1}\left(-\sqrt{3}\right)} is {-\dfrac{π}{3}}
5. Find the value of the following: {\cos^{-1}\left(-\dfrac{1}{2}\right)}
Let {\cos^{-1}\left(-\dfrac{1}{2}\right) = y}. Then,
{\cos y}
=
-\dfrac{1}{2}
=
{\cos\left(π - \dfrac{π}{3}\right)}
=
{\cos\left(\dfrac{3π- π}{3}\right)}
=
{\cos\left(\dfrac{2π}{3}\right)}
We know that the range of principal value branch of cos-1 is [0, π] and {\cos\left(\dfrac{2π}{3}\right) = -\dfrac{1}{2}}.
∴ The principal value of {\cos^{-1}\left(-\dfrac{1}{2}\right)} is {-\dfrac{2π}{3}}
6. Find the value of the following: tan-1(-1)
Let {\tan^{-1}(-1) = y}. Then,
{\tan y}
=
-1
=
{\tan\left(-\dfrac{π}{4}\right)}
We know that the range of principal value branch of tan-1 is {\left(-\dfrac{π}{2}, \dfrac{π}{2}\right)} and {\tan\left(-\dfrac{π}{4}\right) = -1}.
∴ The principal value of tan-1(-1) is {-\dfrac{π}{4}}
7. Find the value of the following: {\sec^{-1}\left(\dfrac{2}{\sqrt{3}}\right)}
Let {\sec^{-1}\left(\dfrac{2}{\sqrt{3}}\right) = y}. Then,
{\sec y}
=
\dfrac{2}{\sqrt{3}}
=
{\sec\left(\dfrac{π}{6}\right)}
We know that the range of principal value branch of sec-1 is {[0, π] - \left\{\dfrac{π}{2}\right\}} and {\sec\left(\dfrac{π}{6}\right) = \dfrac{2}{\sqrt{3}}}.
∴ The principal value of {\sec^{-1}\left(\dfrac{2}{\sqrt{3}}\right)} is {\dfrac{π}{6}}
8. Find the value of the following: {\cot^{-1}\left(\sqrt{3}\right)}
Let {\cot^{-1}\left(\sqrt{3}\right) = y}. Then,
{\cot y}
=
\sqrt{3}
=
{\cot\left(\dfrac{π}{6}\right)}
We know that the range of principal value branch of cot-1 is (0, π) and {\cot\left(\dfrac{π}{6}\right) = \sqrt{3}}.
∴ The principal value of {\cot^{-1}\left(\sqrt{3}\right)} is {\dfrac{π}{6}}
9. Find the value of the following: {\cos^{-1}\left(-\dfrac{1}{\sqrt{2}}\right)}
Let {\cos^{-1}\left(-\dfrac{1}{\sqrt{2}}\right) = y}. Then,
{\cos y}
=
-\dfrac{1}{\sqrt{2}}
=
{\cos\left(π - \dfrac{π}{4}\right)}
=
{\cos\left(\dfrac{4π - π}{4}\right)}
=
{\cos\left(\dfrac{3π}{4}\right)}
We know that the range of principal value branch of cos-1 is {[0, π]} and {\cos\left(\dfrac{3π}{4}\right) = -\dfrac{1}{\sqrt{2}}}.
∴ The principal value of {\cos^{-1}\left(-\dfrac{1}{\sqrt{2}}\right)} is {\dfrac{π}{4}}
10. Find the value of the following: {\cosec^{-1}\left(-\sqrt{2}\right)}
Now, let {\cosec^{-1}\left(-\sqrt{2}\right) = y}. Then,
{\cosec y}
=
-\sqrt{2}
=
{\cosec\left(-\dfrac{π}{4}\right)}
We know that the range of principal value branch of cosec-1 is {\left[-\dfrac{π}{2}, \dfrac{π}{2}\right] - \{0\}} and {\cosec\left(-\dfrac{π}{4}\right) = -\sqrt{2}}.
∴ The principal value of {\cosec^{-1}\left(-\sqrt{2}\right)} is {-\dfrac{π}{4}}
11. Find the values of the following: {\tan^{-1}(1)} {+ \cos^{-1}\left(-\dfrac{1}{2}\right)} {+ \sin^{-1}\left(-\dfrac{1}{2}\right)}
Let {\tan^{-1}(1) = α}. Then,
{\tan α}
=
1
=
{\tan\left(-\dfrac{π}{4}\right)}
We know that the range of principal value branch of tan-1 is {\left(-\dfrac{π}{2}, \dfrac{π}{2}\right)} and {\tan\left(\dfrac{π}{4}\right) = 1}.
∴ The principal value of {\tan^{-1}(1)} is {\dfrac{π}{4}}
Now, let {\cos^{-1}\left(-\dfrac{1}{2}\right) = β}. Then,
{\cos β}
=
{-\dfrac{1}{2}}
=
{\cos\left(π - \dfrac{π}{3}\right)}
=
{\cos\left(\dfrac{3π - π}{3}\right)}
=
{\cos\left(\dfrac{2π}{3}\right)}
We know that the range of principal value branch of cos-1 is {[0, π]} and {\cos\left(\dfrac{2π}{3}\right) = -\dfrac{1}{2}}.
∴ The principal value of {\cos^{-1}\left(-\dfrac{1}{2}\right)} is {\dfrac{2π}{3}}
Let {\sin^{-1}\left(-\dfrac{1}{2}\right) = γ}. Then,
{\sin γ}
=
-\dfrac{1}{2}
=
{\sin\left(-\dfrac{π}{6}\right)}
We know that the range of principal value branch of sin-1 is {\left[-\dfrac{π}{2}, \dfrac{π}{2}\right]} and {\sin\left(-\dfrac{π}{6}\right) = -\dfrac{1}{2}}.
∴ The principal value of {\sin^{-1}\left(-\dfrac{1}{2}\right)} is {-\dfrac{π}{6}}
{∴ \tan^{-1}(1)} {+ \cos^{-1}\left(-\dfrac{1}{2}\right)} {+ \sin^{-1}\left(-\dfrac{1}{2}\right)}
{= \dfrac{π}{4} + \dfrac{2π}{3} - \dfrac{π}{6}}
{= \dfrac{3π + 8π - 2π}{12}}
{= \dfrac{9π}{12}}
= {\dfrac{3π}{4}}
12. Find the values of the following: {\cos^{-1}\left(\dfrac{1}{2}\right) + 2 \sin^{-1}\left(\dfrac{1}{2}\right)}
Let {\cos^{-1}\left(\dfrac{1}{2}\right) = α}. Then,
{\cos α}
=
\dfrac{1}{2}
=
{\cos\left(\dfrac{π}{3}\right)}
We know that the range of principal value branch of cos-1 is {[0, π]} and {\cos\left(\dfrac{π}{3}\right) = \dfrac{1}{2}}.
∴ The principal value of {\cos^{-1}\left(\dfrac{1}{2}\right)} is {\dfrac{π}{3}}
Now, let {\sin^{-1}\left(\dfrac{1}{2}\right) = β}. Then,
{\sin β}
=
\dfrac{1}{2}
=
{\sin\left(\dfrac{π}{6}\right)}
We know that the range of principal value branch of sin-1 is {\left[-\dfrac{π}{2}, \dfrac{π}{2}\right]} and {\sin\left(\dfrac{π}{6}\right) = \dfrac{1}{2}}.
∴ The principal value of {\sin^{-1}\left(\dfrac{1}{2}\right)} is {\dfrac{π}{6}}
{∴ \cos^{-1}\left(\dfrac{1}{2}\right) + 2 \sin^{-1}\left(\dfrac{1}{2}\right)}
{= \dfrac{π}{3} + 2 × \dfrac{π}{6}}
{= \dfrac{π}{3} + \dfrac{π}{3}}
= {\dfrac{2π}{3}}
13. If {\sin^{-1} x = y}, then
(A) {0 \leq y \leq x}
(B) {-\dfrac{π}{2} \leq y \leq \dfrac{π}{2}}
(C) {0 \lt y \lt x}
(D) {-\dfrac{π}{2} \lt y \lt \dfrac{π}{2}}
As it is given that {\sin^{-1} x = y}, it implies that the values of x comprises the domain the values of y comprises the range.
As per the definition, the range (values of y) of principal value branch of sin-1 is {\left[-\dfrac{π}{2}, \dfrac{π}{2}\right]}.
i.e. {-\dfrac{π}{2} \leq y \leq \dfrac{π}{2}}
So, option B is the correct answer.
14. {\tan^{-1}\left(\sqrt{3}\right) - \sec^{-1}(-2)} is equal to
(A) π
(B) {-\dfrac{π}{3}}
(C) {\dfrac{π}{3}}
(D) {\dfrac{2π}{3}}
Let {\tan^{-1}\left(\sqrt{3}\right) = α}. Then,
{\tan α}
=
\sqrt{3}
=
{\tan\left(\dfrac{π}{3}\right)}
We know that the range of principal value branch of tan-1 is {\left(-\dfrac{π}{2}, \dfrac{π}{2}\right)} and {\tan\left(\dfrac{π}{3}\right) = \sqrt{3}}.
∴ The principal value of {\tan^{-1}\left(\sqrt{3}\right)} is {\dfrac{π}{3}}
Now, let {\sec^{-1}(-2) = β}. Then,
{\sec β}
=
-2
=
{\sec\left(π - \dfrac{π}{3}\right)}
=
{\sec\left(\dfrac{3π - π}{3}\right)}
=
{\sec\left(\dfrac{2π}{3}\right)}
We know that the range of principal value branch of sec-1 is {[0, π] - \left\{\dfrac{π}{2}\right\}} and {\sec\left(\dfrac{2π}{3}\right) = -2}.
∴ The principal value of {\sec^{-1}(-2)} is {\dfrac{2π}{3}}
Now, {\tan^{-1}\left(\sqrt{3}\right) - \sec^{-1}(-2)}
{= \dfrac{π}{3} - \dfrac{2π}{3}}
{= \dfrac{2π - π}{3}}
= {-\dfrac{π}{3}}
So, option B is the correct answer.