#### Answer

$\dfrac {\sqrt {3}\tan \theta +1}{\sqrt {3}-\tan \theta }$

#### Work Step by Step

$\tan \left( \theta +30\right) =\dfrac {\tan \theta +\tan 30}{1-\tan \theta \tan 30}=\dfrac {\tan \theta +\dfrac {1}{\sqrt {3}}}{1-\dfrac {1}{\sqrt {3}}\tan \theta }=\dfrac {\sqrt {3}\tan \theta +1}{\sqrt {3}-\tan \theta }$