Answer
$\frac{tan~\frac{9\pi}{8}-tan~\frac{\pi}{8}}{1+tan~\frac{9\pi}{8}~tan~\frac{\pi}{8}}=0$
Work Step by Step
$tan(u-v)=\frac{tan~u-tan~v}{1+tan~u~tan~v}$
$\frac{tan~\frac{9\pi}{8}-tan~\frac{\pi}{8}}{1+tan~\frac{9\pi}{8}~tan~\frac{\pi}{8}}=tan(\frac{9\pi}{8}-\frac{\pi}{8})=tan~\frac{8\pi}{8}=tan~\pi=0$
