Answer
$\dfrac{2 \pi}{3}$
Work Step by Step
We know that the trigonometric function $\cot^{-1} x$ has range: $(0, \pi)$.
Here, $\cot^{-1} (\dfrac{-\sqrt 3}{3})=\dfrac{2 \pi}{ 3}$, because $\sec(\dfrac{2\pi}{3})=\dfrac{-\sqrt 3}{3}$ and $\dfrac{2 \pi}{3}$ lies in the range of $\cot^{-1}{x}$ .