Chapter 4 - Section 4.7 - Inverse Trigonometric Functions - Exercise Set - Page 626: 13

$\dfrac{\pi}{6}$

Suppose $\theta = \tan^{-1} (\dfrac{\sqrt 3}{3})$ This gives: $\tan \theta= \dfrac{\sqrt 3}{3}$ Now, $\theta = \tan^{-1} (\dfrac{\sqrt 3}{3})=\dfrac{\pi}{6}$ Therefore, $\tan^{-1} (\dfrac{\sqrt 3}{3})=\dfrac{\pi}{6}$