已知 $\sin \alpha-\cos \alpha=\sqrt{2}, \alpha \in(0, \pi)$, 则 $\tan \alpha=(\quad)$
由 $\sin \alpha-\cos \alpha=\sqrt{2}(1)$, 两边平方得 $1-2 \sin \alpha \cos \alpha=2$, 即 $2 \sin \alpha \cos \alpha=-1$, 故 $(\sin \alpha+\cos \alpha)^{2}=1$

$+2 \sin \alpha \cos \alpha=0$, 即 $\sin \alpha+\cos \alpha=0$ (2), 联立(1)(2)得 $\sin \alpha=\frac{\sqrt{2}}{2}, \cos \alpha=-\frac{\sqrt{2}}{2}$, 故 $\tan \alpha=\frac{\sin \alpha}{\cos \alpha}=-1$, 故选 A.

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