Question

Find the principal value of ${\cot }^{-1}\left(\sqrt{3}\right)$

Answer 1

Solve using standard trigonometric angle value:

The given inverse trigonometric function is ${\cot }^{-1}\left(\sqrt{3}\right)$.

Let, ${\cot }^{-1}\left(\sqrt{3}\right)=y$

$\Rightarrow$ $\sqrt{3}=\cot y$ $\left[\because {\cot }^{-1}x=y\Rightarrow x=\cot y\right]$

$\Rightarrow$ $\cot y=\sqrt{3}$

$\Rightarrow$ $\cot y=\cot \frac{\mathrm{\pi }}{6}$ $[\because \cot \frac{\pi }{6}=\sqrt{3}]$

$\Rightarrow$ $y=\frac{\mathrm{\pi }}{6}$

As we know, the range of principal value of ${\cot }^{-1}\mathrm{x}$ is $\left(0,\mathrm{\pi }\right)$.

So, the principal value of ${\cot }^{-1}\left(\sqrt{3}\right)$ is $\frac{\mathrm{\pi }}{6}$.

Hence, the required principal value is $\frac{\mathrm{\pi }}{6}$.