Answer
$\dfrac{ \pi }{6}$
Work Step by Step
We know that the trigonometric function $\cot^{-1} x$ has range: $(0, \pi)$.
Here, $\cot^{-1} (\sqrt 3)=\dfrac{\pi}{6}$, because $\cot(\dfrac{\pi}{6})=\dfrac{\sqrt 3}{1}$ and $\dfrac{\pi}{6}$ lies in the range of $\cot^{-1}{x}$ .