# Ch. - 2 Inverse Trigonometric Functions

## Chapter 2 Ex.2.1 Question 1

Find the principal value of $${\sin ^{ - 1}}\left( { - \frac{1}{2}} \right)$$.

### Solution

Let, $${\sin ^{ - 1}}\left( { - \frac{1}{2}} \right) = y$$

Hence,

\begin{align}\sin y &= \left( { - \frac{1}{2}} \right)\\&= - \sin \left( {\frac{\pi }{6}} \right)\\&= \sin \left( { - \frac{\pi }{6}} \right)\end{align}

Range of the principal value of $${\sin ^{ - 1}}\left( x \right)$$ is $$\left( { - \frac{\pi }{2},\frac{\pi }{2}} \right)$$

Thus, principal value of $${\sin ^{ - 1}}\left( { - \frac{1}{2}} \right) = \left( { - \frac{\pi }{6}} \right)$$.

## Chapter 2 Ex.2.1 Question 2

Find the principal value of $${\cos ^{ - 1}}\left( {\frac{{\sqrt 3 }}{2}} \right)$$.

### Solution

Let, $${\cos ^{ - 1}}\left( {\frac{{\sqrt 3 }}{2}} \right) = y$$

Hence,

\begin{align}\cos y &= \left( {\frac{{\sqrt 3 }}{2}} \right)\\&= \cos \frac{\pi }{6}\end{align}

Range of the principal value of $${\cos ^{ - 1}}\left( x \right)$$ is $$\left( {0,\pi } \right)$$.

Thus, principal value of $${\cos ^{ - 1}}\left( {\frac{{\sqrt 3 }}{2}} \right) = \left( {\frac{\pi }{6}} \right)$$

## Chapter 2 Ex.2.1 Question 3

Find the principal value of $$\ {\rm{cosec}}{^{ - 1}}\left( 2 \right)$$.

### Solution

Let, $$\ {\rm{cosec}}^{ - 1}\left( 2 \right) = y$$

Hence,

\begin{align}\ {\rm{cosec}}\,y &= 2\\ &= {\rm{cosec}}\left( {\frac{\pi }{6}} \right)\end{align}

Range of the principal value of $$\ {\rm{cosec}}{^{ - 1}}\left( x \right) = \left[ { - \frac{\pi }{2},\frac{\pi }{2}} \right] - \left\{ 0 \right\}$$

Thus, principal value of $$\ {\rm{cosec}}{^{ - 1}}\left( 2 \right) = \left( {\frac{\pi }{6}} \right)$$.

## Chapter 2 Ex.2.1 Question 4

Find the principal value of $${\tan ^{ - 1}}\left( { - \sqrt 3 } \right)$$

### Solution

Let, $${\tan ^{ - 1}}\left( { - \sqrt 3 } \right) = y$$

Hence,

\begin{align}\tan y &= - \sqrt 3 \\&= - \tan \left( {\frac{\pi }{3}} \right)\\&= \tan \left( { - \frac{\pi }{3}} \right)\end{align}

Range of the principal value of $${\tan ^{ - 1}}\left( x \right) = \left( { - \frac{\pi }{2},\frac{\pi }{2}} \right)$$

Thus, principal value of $${\tan ^{ - 1}}\left( { - \sqrt 3 } \right) = \left( { - \frac{\pi }{3}} \right)$$.

## Chapter 2 Ex.2.1 Question 5

Find the principal value of $${\cos ^{ - 1}}\left( { - \frac{1}{2}} \right)$$

### Solution

Let, $${\cos ^{ - 1}}\left( { - \frac{1}{2}} \right) = y$$

Hence,

\begin{align}\cos y &= - \frac{1}{2}\\&= - \cos \left( {\frac{\pi }{3}} \right)\\&= \cos \left( {\pi - \frac{\pi }{3}} \right)\\&= \cos \left( {\frac{{2\pi }}{3}} \right)\end{align}

Range of the principal value of $${\cos ^{ - 1}}\left( x \right) = \left[ {0,\pi } \right]$$

Thus, principal value of $${\cos ^{ - 1}}\left( { - \frac{1}{2}} \right) = \left( {\frac{{2\pi }}{3}} \right)$$.

## Chapter 2 Ex.2.1 Question 6

Find the principal value of $${\tan ^{ - 1}}\left( { - 1} \right)$$

### Solution

Let, $${\tan ^{ - 1}}\left( { - 1} \right) = y$$

Hence,

\begin{align}\tan {\rm{ }}y &= - 1\\&= - \tan \left( {\frac{\pi }{4}} \right)\\&= \tan \left( { - \frac{\pi }{4}} \right)\end{align}

Range of the principal value of $${\tan ^{ - 1}}\left( x \right) = \left( { - \frac{\pi }{2},\frac{\pi }{2}} \right)$$

Thus, principal value of $${\tan ^{ - 1}}\left( { - 1} \right) = \left( { - \frac{\pi }{4}} \right)$$.

## Chapter 2 Ex.2.1 Question 7

Find the principal value of $${\sec ^{ - 1}}\left( {\frac{2}{{\sqrt 3 }}} \right)$$

### Solution

Let, $${\sec ^{ - 1}}\left( {\frac{2}{{\sqrt 3 }}} \right)=y$$

Hence,

\begin{align} \sec \;y &= \frac{2}{{\sqrt 3 }} \\ &= \sec \left( {\frac{\pi }{6}} \right) \end{align}

Range of the principal value of $${\sec ^{ - 1}}\left( x \right) = \left[ {0,\pi } \right] - \left\{ {\frac{\pi }{2}} \right\}$$

Thus, principal value of $${\sec ^{ - 1}}\left( {\frac{2}{{\sqrt 3 }}} \right) = \left( {\frac{\pi }{6}} \right)$$.

## Chapter 2 Ex.2.1 Question 8

Find the principal value of $${\cot ^{ - 1}}\left( {\sqrt 3 } \right)$$

### Solution

Let, $${\cot ^{ - 1}}\left( {\sqrt 3 } \right) = y$$

Hence,

\begin{align}\cot \;y &= \sqrt 3 \\&= \cot \left( {\frac{\pi }{6}} \right)\end{align}

Range of the principal value of $${\cot ^{ - 1}}\left( x \right) = \left( {0,\pi } \right)$$

Thus, principal value of $${\cot ^{ - 1}}\left( {\sqrt 3 } \right) = \left( {\frac{\pi }{6}} \right)$$.

## Chapter 2 Ex.2.1 Question 9

Find the principal value of $${\cos ^{ - 1}}\left( { - \frac{1}{{\sqrt 2 }}} \right)$$

### Solution

Let, $${\cos ^{ - 1}}\left( { - \frac{1}{{\sqrt 2 }}} \right) = y$$

Hence,

\begin{align}\cos \;y &= - \frac{1}{{\sqrt 2 }}\\&= - \cos \left( {\frac{\pi }{4}} \right)\\&= \cos \left( { - \frac{\pi }{4}} \right)\\&= \cos \left( {\pi - \frac{\pi }{4}} \right)\\&= \cos \left( {\frac{{3\pi }}{4}} \right)\end{align}

Range of the principal value of $${\cos ^{ - 1}}\left( x \right) = \left[ {0,\pi } \right]$$

Thus, principal value of $${\cos ^{ - 1}}\left( { - \frac{1}{{\sqrt 2 }}} \right) = \left( {\frac{{3\pi }}{4}} \right)$$.

## Chapter 2 Ex.2.1 Question 10

Find the principal value of $${\rm{cose}}{{\rm{c}}^{ - 1}}\left( { - \sqrt 2 } \right)$$

### Solution

Let, $${\rm{cose}}{{\rm{c}}^{ - 1}}\left( { - \sqrt 2 } \right) = y$$

Hence,

\begin{align}{\rm{cosec }}\;y &= - \sqrt 2 \\&= - {\rm{cosec}}\left( {\frac{\pi }{4}} \right)\\&= {\rm{cosec}}\left( { - \frac{\pi }{4}} \right)\end{align}

Range of the principal value of $${\rm{cose}}{{\rm{c}}^{ - 1}}\left( x \right) = \left[ { - \frac{\pi }{2},\frac{\pi }{2}} \right] - \left\{ 0 \right\}$$

Thus, principal value of $${\rm{cose}}{{\rm{c}}^{ - 1}}\left( { - \sqrt 2 } \right) = \left( { - \frac{\pi }{4}} \right)$$.

## Chapter 2 Ex.2.1 Question 11

Find the value of $${\tan ^{ - 1}}\left( 1 \right) + {\cos ^{ - 1}}\left( { - \frac{1}{2}} \right) + {\sin ^{ - 1}}\left( { - \frac{1}{2}} \right)$$.

### Solution

Let, $${\tan ^{ - 1}}\left( 1 \right) = x$$

Hence,

\begin{align}\tan x &= 1\\&= \tan \left( {\frac{\pi }{4}} \right)\end{align}

Therefore,

$${\tan ^{ - 1}}\left( 1 \right) = \left( {\frac{\pi }{4}} \right)$$

Now, let $${\cos ^{ - 1}}\left( { - \frac{1}{2}} \right) = y$$

Hence,

\begin{align}\cos y &= - \frac{1}{2}\\&= - \cos \left( {\frac{\pi }{3}} \right)\\&= \cos \left( {\pi - \frac{\pi }{3}} \right)\\&= \cos \left( {\frac{{2\pi }}{3}} \right)\end{align}

Therefore,

$${\cos ^{ - 1}}\left( { - \frac{1}{2}} \right) = \frac{{2\pi }}{3}$$

Again, let $${\sin ^{ - 1}}\left( { - \frac{1}{2}} \right) = z$$

Hence,

\begin{align}\sin z &= - \frac{1}{2}\\&= - \sin \left( {\frac{\pi }{6}} \right)\\&= \sin \left( { - \frac{\pi }{6}} \right)\end{align}

Therefore,

$${\sin ^{ - 1}}\left( { - \frac{1}{2}} \right) = - \frac{\pi }{6}$$

Thus,

\begin{align}{\tan ^{ - 1}}\left( 1 \right) + {\cos ^{ - 1}}\left( { - \frac{1}{2}} \right) + {\sin ^{ - 1}}\left( { - \frac{1}{2}} \right) &= \frac{\pi }{4} + \frac{{2\pi }}{3} - \frac{\pi }{6}\\&= \frac{{3\pi + 8\pi - 2\pi }}{{12}}\\&= \frac{{9\pi }}{{12}}\\&= \frac{{3\pi }}{4}\end{align}

## Chapter 2 Ex.2.1 Question 12

Find the value of $${\cos ^{ - 1}}\left( {\frac{1}{2}} \right) + 2{\sin ^{ - 1}}\left( {\frac{1}{2}} \right)$$

### Solution

Let, $${\tan ^{ - 1}}\left( 1 \right) = x$$

Hence,

\begin{align}\cos x &= \frac{1}{2}\\&= \cos \left( {\frac{\pi }{3}} \right)\end{align}

Therefore,

$${\cos ^{ - 1}}\left( {\frac{1}{2}} \right) = \frac{\pi }{3}$$

Let, $${\sin ^{ - 1}}\left( {\frac{1}{2}} \right) = y$$

Hence,

\begin{align}\sin y &= \frac{1}{2}\\&= \sin \left( {\frac{\pi }{6}} \right)\end{align}

Therefore,

$${\sin ^{ - 1}}\left( {\frac{1}{2}} \right) = \frac{\pi }{6}$$

Thus

\begin{align}{\cos ^{ - 1}}\left( {\frac{1}{2}} \right) + 2{\sin ^{ - 1}}\left( {\frac{1}{2}} \right) &= \frac{\pi }{3} + 2\left( {\frac{\pi }{6}} \right)\\&= \frac{{2\pi }}{3}\end{align}

## Chapter 2 Ex.2.1 Question 13

Find the value of $${\sin ^{ - 1}}x = y$$, then

(A) $$0 \le y \le \pi$$

(B) $$- \frac{\pi }{2} \le y \le \frac{\pi }{2}$$

(C) $$0 \le y \le \pi$$

(D) $$- \frac{\pi }{2} < y < \frac{\pi }{2}$$

### Solution

It is given that $${\sin ^{ - 1}}x = y$$

Range of the principal value of $${\sin ^{ - 1}}x = \left[ { - \frac{\pi }{2},\frac{\pi }{2}} \right]$$

Thus, $$- \frac{\pi }{2} \le y \le \frac{\pi }{2}$$

## Chapter 2 Ex.2.1 Question 14

Find the value of $${\tan ^{ - 1}}\sqrt 3 - {\sec ^{ - 1}}\left( { - 2} \right)$$is equal to

(A) $$0$$

(B) $$- \frac{\pi }{3}$$

(C) $$\frac{\pi }{3}$$

(D) $$\frac{{2\pi }}{3}$$

### Solution

Let

Hence,

\begin{align}\tan x &= \sqrt 3 \\&= \tan \left( {\frac{\pi }{3}} \right)\end{align}

Range of the principal value of $${\tan ^{ - 1}}x = \left( { - \frac{\pi }{2},\frac{\pi }{2}} \right)$$

Therefore,$${\tan ^{ - 1}}\left( {\sqrt 3 } \right) = \left( {\frac{\pi }{3}} \right)$$

Let $${\sec ^{ - 1}}\left( { - 2} \right) = y$$

Hence,

\begin{align}\sec y &= \left( { - 2} \right)\\&= - \sec \left( {\frac{\pi }{3}} \right)\\&= \sec \left( { - \frac{\pi }{3}} \right)\\&= \sec \left( {\pi - \frac{\pi }{3}} \right)\\&= \sec \left( {\frac{{2\pi }}{3}} \right)\end{align}

Range of the principal value of $${\rm{se}}{{\rm{c}}^{ - 1}}{\rm{x = }}\left[ {0,\pi } \right] - \left\{ {\frac{\pi }{2}} \right\}$$

Therefore, $${\sec ^{ - 1}}\left( { - 2} \right) = \frac{{2\pi }}{3}$$

Thus,

\begin{align}{\tan ^{ - 1}}\sqrt 3 - {\sec ^{ - 1}}\left( { - 2} \right)& = \frac{\pi }{3} - \frac{{2\pi }}{3}\\&= - \frac{\pi }{3}\end{align}

The answer is $$B$$.

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