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 $\cos ^{-1}\left(\frac{\sqrt{3}}{2}\right)=\left(\frac{\pi}{6}\right)$

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 $\operatorname{cosec}^{-1}(2)=\left(\frac{\pi}{6}ight)$.

2. Inverse Trigonometric Functions

2.1

2.2

Miscellaneous