Chapter 6 - Analytic Trigonometry - Section 6.2 The Inverse Trigonometric Functions (Continued) - 6.2 Assess Your Understanding - Page 480: 22

$\dfrac{-\pi}{6}$

The range of $\tan^{-1}{x}$ is $[-\dfrac{ \pi}{2}, \dfrac{\pi}{2}]$. Here, $\cot(\dfrac{2\pi}{3})=\dfrac{ -\sqrt 3}{3}$, because $\tan{(\dfrac{-\pi}{6})}=\dfrac{-\sqrt 3}{3}$ and $\dfrac{-\pi}{6}$ lies in the range of $\tan^{-1}{x}$ . Therefore, we have: $\tan^{-1}(\cot {(\dfrac{2\pi }{3})}]=\tan^{-1} ({\dfrac{- \sqrt 3}{3}})=\dfrac{-\pi}{6}$