Answer
$ \dfrac{\pi}{6}$
Work Step by Step
$y = \tan^{-1}{x} \hspace{15pt} \to \hspace{15pt} x = \tan{y}$
$\text{where } \hspace{15pt} -\infty \leq x \leq \infty \hspace{15pt} \text{and} \hspace{15pt} -\dfrac{\pi}{2} \leq y \leq \dfrac{\pi}{2}$
Let $\theta = \tan^{-1} {\left(\dfrac{\sqrt{3}}{3} \right)} \hspace{20pt} -\dfrac{\pi}{2} \leq \theta \leq \dfrac{\pi}{2}$
Since $\theta = \tan^{-1}{\left(\dfrac{\sqrt{3}}{3} \right)}$, then
$\tan{\theta} =\dfrac{\sqrt{3}}{3}$
Referring to Table 3 on page 472:
The only angle in the interval $\left[-\dfrac{\pi}{2},\dfrac{\pi}{2} \right]$ whose tangent is $\dfrac{\sqrt{3}}{3}$ is $\dfrac{\pi}{6}$.
Therefore, $\tan^{-1}{\left(\dfrac{\sqrt{3}}{3} \right)} = \dfrac{\pi}{6}$.
