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Q.
The principal value of $\sec ^{-1} \frac{2}{\sqrt{3}}$ is ..$A$.. Here, $A$ refers to
Inverse Trigonometric Functions
Solution:
Let $\sec ^{-1}\left(\frac{2}{\sqrt{3}}\right)=\theta \Rightarrow \sec \theta=\frac{2}{\sqrt{3}}$
We know that, the range of principal value branch of $\sec ^{-1} \theta$ is $[0, \pi]-\left\{\frac{\pi}{2}\right\}$.
$\because \sec \theta=\frac{2}{\sqrt{3}}=\sec \frac{\pi}{6} \Rightarrow \theta=\frac{\pi}{6}$ where, $\theta \in[0, \pi]-\left\{\frac{\pi}{2}\right\}$
$\Rightarrow \sec ^{-1}\left(\frac{2}{\sqrt{3}}\right)=\frac{\pi}{6}$
Hence, the principal value of $\sec ^{-1}\left(\frac{2}{\sqrt{3}}\right)$ is $\frac{\pi}{6}$.