## Precalculus (10th Edition)

$-\dfrac{\pi}{6}$
Note that $\cot{\left(\dfrac{2\pi}{3}\right)}=-\dfrac{\sqrt3}{3}$. Thus $\tan^{-1}\left(\cot{\left(\dfrac{2\pi}{3}\right)}\right)=\tan^{-1}{\left(-\dfrac{\sqrt3}{3}\right)}=-\dfrac{\pi}{6}$, because $\tan{\left(-\dfrac{\pi}{6}\right)}=-\dfrac{\sqrt3}{3}$ and $-\dfrac{\pi}{6}$ is in the range of $\tan^{-1}{x}$, which is $\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right].$