Find the principal value of cot^-1 ≤ft( (-1)/(√3) ).

The correct answer is Option (3) → $\frac{2\pi}{3}$ ##

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

$\cot y = \frac{-1}{\sqrt{3}} = -\cot \left( \frac{\pi}{3} \right) = \cot \left( \pi - \frac{\pi}{3} \right) = \cot \left( \frac{2\pi}{3} \right)$

We know that the range of principal value branch of $\cot^{-1}$ is $(0, \pi)$ and $\cot \left( \frac{2\pi}{3} \right) = \frac{-1}{\sqrt{3}}$.

Hence, principal value of $\cot^{-1} \left( \frac{-1}{\sqrt{3}} \right)$ is $\frac{2\pi}{3}$.