Question
The principal value of $\cot ^{-1}(-\sqrt{3})$ is
✓
Answer
We know that $\cot ^{-1}(x) \in(0, \pi)$
$\begin{array}{ll}
\cot ^{-1}(-\sqrt{3})=\cot ^{-1}\left(-\cot \frac{\pi}{6}\right) & \\
=\cot ^{-1}\left[\cot \left(\pi-\frac{\pi}{6}\right)\right] & {[\because \cot (\pi-\theta)=-\cot \theta]} \\
=\cot ^{-1}\left[\cot \left(\frac{5 \pi}{6}\right)\right]=\frac{5 \pi}{6} & {\left[\because \cot ^{-1}[\cot \theta]=\theta\right]}
\end{array}
$Thus, the principal value of $\cot ^{-1}(-\sqrt{3})$ is $\frac{5 \pi}{6}$.
$\begin{array}{ll}
\cot ^{-1}(-\sqrt{3})=\cot ^{-1}\left(-\cot \frac{\pi}{6}\right) & \\
=\cot ^{-1}\left[\cot \left(\pi-\frac{\pi}{6}\right)\right] & {[\because \cot (\pi-\theta)=-\cot \theta]} \\
=\cot ^{-1}\left[\cot \left(\frac{5 \pi}{6}\right)\right]=\frac{5 \pi}{6} & {\left[\because \cot ^{-1}[\cot \theta]=\theta\right]}
\end{array}
$Thus, the principal value of $\cot ^{-1}(-\sqrt{3})$ is $\frac{5 \pi}{6}$.
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