Answer
$\dfrac{\sqrt{3}}{3}$
Work Step by Step
$\tan{(\theta+ \pi k}) = tan{\theta}$
$\therefore \tan{\dfrac{19 \pi}{6 }} = \tan{\left(\dfrac{\pi}{6}+ (3 \times \pi) \right)} = \tan{\dfrac{\pi}{6}}$
$\tan{\dfrac{\pi}{6}} =\boxed{\dfrac{\sqrt{3}}{3}}$
