Answer
$\frac{tan~\frac{\pi}{15}+tan~\frac{2\pi}{5}}{1-tan~\frac{\pi}{15}~tan~\frac{2\pi}{5}}=tan(\frac{7\pi}{15})$
Work Step by Step
$tan(u+v)=\frac{tan~u+tan~v}{1-tan~u~tan~v}$
$\frac{tan~\frac{\pi}{15}+tan~\frac{2\pi}{5}}{1-tan~\frac{\pi}{15}~tan~\frac{2\pi}{5}}=tan(\frac{\pi}{15}+\frac{2\pi}{5})=tan(\frac{\pi}{15}+\frac{2\pi(3)}{5(3)})=tan(\frac{\pi}{15}+\frac{6\pi}{15})=tan(\frac{7\pi}{15})$
