若 $\frac{\sin \alpha+\cos \alpha}{\sin \alpha-\cos \alpha}=\frac{1}{2}$, 则 $\tan 2 \alpha$ 等于 ( )
$\frac{\sin \alpha+\cos \alpha}{\sin \alpha-\cos \alpha}=\frac{\tan \alpha+1}{\tan \alpha-1}=\frac{1}{2}, \tan \alpha=-3, \tan 2 \alpha=\frac{2 \tan \alpha}{1-\tan ^{2} \alpha}=\frac{-6}{-8}=\frac{3}{4}$.