Question
Find the principal value of $\cot ^{-1}(\sqrt{3})$.
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Answer
$\cot^{-1}x$ represents an angle in $(0, \pi)$ whose cotangent is $x.$
Let $x=\cot ^{-1}(\sqrt{3})$
$\Rightarrow \cot x=\sqrt{3}=\cot \left(\frac{\pi}{6}\right)$
$\Rightarrow x=\frac{\pi}{6}$
$\therefore$ Principal value of $\cot ^{-1}(\sqrt{3})$ is $\frac{\pi}{6}.$
Let $x=\cot ^{-1}(\sqrt{3})$
$\Rightarrow \cot x=\sqrt{3}=\cot \left(\frac{\pi}{6}\right)$
$\Rightarrow x=\frac{\pi}{6}$
$\therefore$ Principal value of $\cot ^{-1}(\sqrt{3})$ is $\frac{\pi}{6}.$
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