已知 $\frac{\sin \alpha+\cos \alpha}{\sin \alpha-\cos \alpha}=2$, 则 $\sin \alpha \cos \alpha$ 的值为 \$ \qquad \$ .
由 $\frac{\sin \alpha+\cos \alpha}{\sin \alpha-\cos \alpha}=2$, 得 $\frac{\tan \alpha+1}{\tan \alpha-1}=2, \therefore \tan \alpha=3, \therefore \sin \alpha \cos \alpha=\frac{\sin \alpha \cos \alpha}{\sin ^{2} \alpha+\cos ^{2} \alpha}=\frac{\tan \alpha}{\tan ^{2} \alpha+1}=\frac{3}{10}$.