## Precalculus (10th Edition)

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