# How do you express tan((19pi)/5) as a trig function of an angle in Quadrant I?

$- \tan \left(\frac{\pi}{5}\right)$
$\tan \left(\frac{19 \pi}{5}\right) = \tan \left(- \frac{\pi}{5} + 20 \frac{\pi}{5}\right) = \tan \left(- \frac{\pi}{5} + 4 \pi\right) =$
$= \tan \left(- \frac{\pi}{5}\right) = - \tan \left(\frac{\pi}{5}\right)$
$\frac{\pi}{5}$ is an arc (or angle) in Quadrant 1.