7.若 $\frac{\sin \alpha+\cos \alpha}{\sin \alpha-\cos \alpha}=\frac{1}{2}$, 则 $\tan 2 \alpha$ 等于
由 $\frac{\sin \alpha+\cos \alpha}{\sin \alpha-\cos \alpha}=\frac{1}{2}$,

得 $2(\sin \alpha+\cos \alpha)=\sin \alpha-\cos \alpha$,

即 $\tan \alpha=-3$.

又 $\tan 2 \alpha=\frac{2 \tan \alpha}{1-\tan ^{2} \alpha}=\frac{-6}{1-9}=\frac{6}{8}=\frac{3}{4}$.
答案: $\frac{3}{4}$