^ -1 (2 3 ) - Vidyadip

Question

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

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Answer

Let $ \sec ^{-1}\left(\frac{2}{\sqrt{3}}\right)=y$
$\Rightarrow \sec y=\frac{2}{\sqrt{3}}$
$\Rightarrow \sec y=\sec \frac{\pi}{6}$
Since, the principal value branch of $\sec ^{-1}$ is $(0, \pi)$.
Therefore, Principal value of $\sec ^{-1}\left(\frac{2}{\sqrt{3}}\right)$ is $\frac{\pi}{6}$.

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