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**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●

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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.