cot^(-1) (-sqrt3)

$$\\cot^{-1}(-\sqrt3)=tan^{-1}\left(\frac{-1}{\sqrt3}\right)\\\\
\mbox{ The first quadrant answer is } \pi/6\quad \mbox{ But the first quadrant in not correct.}\\\\$$

this is the range of the function.

so

answer is     $$-\pi/6$$

Thanks