Answer
$\dfrac {\sqrt {3}}{3}$
Work Step by Step
$\tan \alpha =\tan \left( \alpha -\pi k\right)
\\\Rightarrow \tan \dfrac {19\pi }{6}=\tan \left( \dfrac {19\pi }{6}-3\pi \right)
\\=\tan \dfrac {\pi }{6}
\\=\dfrac {\sin \frac {\pi }{6}}{\cos \frac {\pi }{6}}
\\=\dfrac {\frac {1}{2}}{\frac {\sqrt {3}}{2}}
\\=\dfrac {\sqrt {3}}{3}$
