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

Let, $\tan ^{-1}(-\sqrt{3})=y$ Hence, $$ \begin{aligned} \tan y & =-\sqrt{3} \\ & =-\tan \left(\frac{\pi}{3}\right) \\ & =\tan \left(-\frac{\pi}{3}\right) \end{aligned} $$ Range of the principal value of $\tan ^{-1}(x)=\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ Thus, principal value of $\tan ^{-1}(-\sqrt{3})=\left(-\frac{\pi}{3}\right)$.

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

Let, $\cos ^{-1}\left(-\frac{1}{2}\right)=y$ Hence, $$ \begin{aligned} \cos y & =-\frac{1}{2} \\ & =-\cos \left(\frac{\pi}{3}\right) \\ & =\cos \left(\pi-\frac{\pi}{3}\right) \\ & =\cos \left(\frac{2 \pi}{3}\right) \end{aligned} $$ Range of the principal value of $\cos ^{-1}(x)=[0, \pi]$ Thus, principal value of $\cos ^{-1}\left(-\frac{1}{2}\right)=\left(\frac{2 \pi}{3}\right)$.

Q: Find the principal value of $\tan ^{-1}(-1)$

Let, $\tan ^{-1}(-1)=y$ Hence, $$ \begin{aligned} \tan y & =-1 \\ & =-\tan \left(\frac{\pi}{4}\right) \\ & =\tan \left(-\frac{\pi}{4}\right) \end{aligned} $$ Range of the principal value of $\tan ^{-1}(x)=\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ Thus, principal value of $\tan ^{-1}(-1)=\left(-\frac{\pi}{4}\right)$.

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

Let, $\sec ^{-1}\left(\frac{2}{\sqrt{3}}\right)=y$ Hence, $$ \begin{aligned} \sec y & =\frac{2}{\sqrt{3}} \\ & =\sec \left(\frac{\pi}{6}\right) \end{aligned} $$ Range of the principal value of $\sec ^{-1}(x)=[0, \pi]-\left\{\frac{\pi}{2}\right\}$ Thus, principal value of $\sec ^{-1}\left(\frac{2}{\sqrt{3}}\right)=\left(\frac{\pi}{6}\right)$.

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

Let, $\sin ^{-1}\left(-\frac{1}{2}\right)=y$ Hence, $$ \begin{aligned} \sin y & =\left(-\frac{1}{2}\right) \\ & =-\sin \left(\frac{\pi}{6}\right) \\ & =\sin \left(-\frac{\pi}{6}\right) \end{aligned} $$ Range of the principal value of $\sin ^{-1}(x)$ is $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ Thus, principal value of $\sin ^{-1}\left(-\frac{1}{2}\right)=\left(-\frac{\pi}{6}\right)$.

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

Let, $\cos ^{-1}\left(\frac{\sqrt{3}}{2}\right)=y$ Hence, $$ \begin{aligned} \cos y & =\left(\frac{\sqrt{3}}{2}\right) \\ & =\cos \frac{\pi}{6} \end{aligned} $$ Range of the principal value of $\cos ^{-1}(x)$ is $(0, \pi)$. Thus, principal value of $\cos ^{-1}\left(\frac{\sqrt{3}}{2}\right)=\left(\frac{\pi}{6}\right)$

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

Let, $\cos ^{-1}\left(-\frac{1}{\sqrt{2}}\right)=y$ Hence, $$ \begin{aligned} \cos y & =-\frac{1}{\sqrt{2}} \\ & =-\cos \left(\frac{\pi}{4}\right) \\ & =\cos \left(-\frac{\pi}{4}\right) \\ & =\cos \left(\pi-\frac{\pi}{4}\right) \\ & =\cos \left(\frac{3 \pi}{4}\right) \end{aligned} $$ Range of the principal value of $\cos ^{-1}(x)=[0, \pi]$ Thus, principal value of $\cos ^{-1}\left(-\frac{1}{\sqrt{2}}\right)=\left(\frac{3 \pi}{4}\right)$.

Question 1

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

Accepted Answer

Let, $	an ^{-1}(-\sqrt{3})=y$

Hence,

$$

\begin{aligned}

an y & =-\sqrt{3} \

& =-	an \left(\frac{\pi}{3}ight) \

& =	an \left(-\frac{\pi}{3}ight)

\end{aligned}

$$

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

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

Question 2

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

Accepted Answer

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

Hence,

$$

\begin{aligned}

\cos y & =-\frac{1}{2} \

& =-\cos \left(\frac{\pi}{3}ight) \

& =\cos \left(\pi-\frac{\pi}{3}ight) \

& =\cos \left(\frac{2 \pi}{3}ight)

\end{aligned}

$$

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

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

Question 3

Find the principal value of $	an ^{-1}(-1)$

Accepted Answer

Let, $	an ^{-1}(-1)=y$

Hence,

$$

\begin{aligned}

an y & =-1 \

& =-	an \left(\frac{\pi}{4}ight) \

& =	an \left(-\frac{\pi}{4}ight)

\end{aligned}

$$

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

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

Question 4

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

Accepted Answer

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

Hence,

$$

\begin{aligned}

\sec y & =\frac{2}{\sqrt{3}} \

& =\sec \left(\frac{\pi}{6}ight)

\end{aligned}

$$

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

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

Question 5

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

Accepted Answer

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

Hence,

$$

\begin{aligned}

\cot y & =\sqrt{3} \

& =\cot \left(\frac{\pi}{6}ight)

\end{aligned}

$$

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

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

Question 6

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

Accepted Answer

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

Hence,

$$

\begin{aligned}

\sin y & =\left(-\frac{1}{2}ight) \

& =-\sin \left(\frac{\pi}{6}ight) \

& =\sin \left(-\frac{\pi}{6}ight)

\end{aligned}

$$

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

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

Question 7

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

Accepted Answer

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

Hence,

$$

\begin{aligned}

\cos y & =\left(\frac{\sqrt{3}}{2}ight) \

& =\cos \frac{\pi}{6}

\end{aligned}

$$

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

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

Question 8

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

Accepted Answer

Let, $\operatorname{cosec}^{-1}(2)=y$

Hence,

$$

\begin{aligned}

\operatorname{cosec} y & =2 \

& =\operatorname{cosec}\left(\frac{\pi}{6}ight)

\end{aligned}

$$

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

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

Question 9

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

Accepted Answer

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

Hence,

$$

\begin{aligned}

\cos y & =-\frac{1}{\sqrt{2}} \

& =-\cos \left(\frac{\pi}{4}ight) \

& =\cos \left(-\frac{\pi}{4}ight) \

& =\cos \left(\pi-\frac{\pi}{4}ight) \

& =\cos \left(\frac{3 \pi}{4}ight)

\end{aligned}

$$

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

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

Question 10

Find the principal value of $\operatorname{cosec}^{-1}(-\sqrt{2})$

Accepted Answer

Let, $\operatorname{cosec}^{-1}(-\sqrt{2})=y$

Hence,

$$

\begin{aligned}

\operatorname{cosec} y & =-\sqrt{2} \

& =-\operatorname{cosec}\left(\frac{\pi}{4}ight) \

& =\operatorname{cosec}\left(-\frac{\pi}{4}ight)

\end{aligned}

$$

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

Thus, principal value of $\operatorname{cosec}^{-1}(-\sqrt{2})=\left(-\frac{\pi}{4}ight)$.

Question 11

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

Accepted Answer

Let, $	an ^{-1}(1)=x$

Hence,

$$

\begin{aligned}

an x & =1 \

& =	an \left(\frac{\pi}{4}ight)

\end{aligned}

$$

Therefore,

$$

an ^{-1}(1)=\left(\frac{\pi}{4}ight)

$$

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

Hence,

$$

\begin{aligned}

\cos y & =-\frac{1}{2} \

& =-\cos \left(\frac{\pi}{3}ight) \

& =\cos \left(\pi-\frac{\pi}{3}ight) \

& =\cos \left(\frac{2 \pi}{3}ight)

\end{aligned}

$$

Therefore,

$$

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

$$

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

Hence,

$$

\begin{aligned}

\sin z & =-\frac{1}{2} \

& =-\sin \left(\frac{\pi}{6}ight) \

& =\sin \left(-\frac{\pi}{6}ight)

\end{aligned}

$$

Therefore,

$$

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

$$

Thus,

$$

\begin{aligned}

an ^{-1}(1)+\cos ^{-1}\left(-\frac{1}{2}ight)+\sin ^{-1}\left(-\frac{1}{2}ight) & =\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{aligned}

$$

Question 12

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

Accepted Answer

Let, $	an ^{-1}(1)=x$

Hence,

$$

\begin{aligned}

\cos x & =\frac{1}{2} \

& =\cos \left(\frac{\pi}{3}ight)

\end{aligned}

$$

Therefore,

$$

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

$$

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

Hence,

$$

\begin{aligned}

\sin y & =\frac{1}{2} \

& =\sin \left(\frac{\pi}{6}ight)

\end{aligned}

$$

Therefore,

$$

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

$$

Thus

$$

\begin{aligned}

\cos ^{-1}\left(\frac{1}{2}ight)+2 \sin ^{-1}\left(\frac{1}{2}ight) & =\frac{\pi}{3}+2\left(\frac{\pi}{6}ight) \

& =\frac{2 \pi}{3}

\end{aligned}

$$

Question 13

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

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

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

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

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

Accepted Answer

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, $-\frac{\pi}{2} \leq y \leq \frac{\pi}{2}$

The answer is $B$.

Question 14

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

(A) 0

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

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

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

Accepted Answer

Let $	an ^{-1}(\sqrt{3})=x$

Hence,

$$

\begin{aligned}

an x & =\sqrt{3} \

& =	an \left(\frac{\pi}{3}ight)

\end{aligned}

$$

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

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

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

Hence,

$$

\begin{aligned}

\sec y & =(-2) \

& =-\sec \left(\frac{\pi}{3}ight) \

& =\sec \left(-\frac{\pi}{3}ight) \

& =\sec \left(\pi-\frac{\pi}{3}ight) \

& =\sec \left(\frac{2 \pi}{3}ight)

\end{aligned}

$$

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

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

Thus,

$$

\begin{aligned}

an ^{-1} \sqrt{3}-\sec ^{-1}(-2) & =\frac{\pi}{3}-\frac{2 \pi}{3} \

& =-\frac{\pi}{3}

\end{aligned}

$$

The answer is $B$.

\section*{EXERCISE 2.2}