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

Answer

(c) : Let $\sec ^{-1}\left(\frac{-2}{\sqrt{3}}\right)=\theta \Rightarrow \sec \theta=\frac{-2}{\sqrt{3}}=-\sec \frac{\pi}{6}$
$
\begin{aligned}
& =\sec \left(\pi-\frac{\pi}{6}\right)=\sec \frac{5 \pi}{6} \\
\Rightarrow & \theta=\frac{5 \pi}{6} \in[0, \pi]-\left\{\frac{\pi}{2}\right\}
\end{aligned}
$
$\therefore \quad$ Principal value of $\sec ^{-1}\left(\frac{-2}{\sqrt{3}}\right)$ is $\frac{5 \pi}{6}$.

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