Answer
$\dfrac{\pi}{6}$
Work Step by Step
Consider the Maclaurin Series for $ \tan^{-1} x$ as follows:
$ \tan^{-1} x=\Sigma_{n=0}^\infty (-1)^n\dfrac{x^{2n-1}}{(2n-1)}=x-\dfrac{x^3}{3}+\dfrac{x^5}{5}-\dfrac{x^7}{7}+...$
As we are given that $x=\dfrac{1}{\sqrt 3}$
Now, we have
$\tan^{-1} x= \tan^{-1} (\dfrac{1}{\sqrt 3})$
$\implies \tan^{-1} (\tan (\dfrac{\pi}{6}))=\dfrac{\pi}{6}$
