若 $\sin \alpha+\cos \alpha=\frac{1}{3}, \alpha \in(0, \pi)$, 则 $\frac{1+\tan \alpha}{1-\tan \alpha}=（ ）$

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由 $\sin \alpha+\cos \alpha=\frac{1}{3}$, 可得 $(\sin \alpha+\cos \alpha)^{2}=1+2 \sin \alpha \cos \alpha=\frac{1}{9}$,

解得 $2 \sin \alpha \cos \alpha=-\frac{8}{9}<0$, 即 $\sin \alpha$ 与 $\cos \alpha$ 异号,

又因为 $\alpha \in(0, \pi)$, 所以 $\sin \alpha>0, \cos \alpha<0$,

又由 $(\sin \alpha-\cos \alpha)^{2}=1-2 \sin \alpha \cos \alpha=\frac{17}{9}$, 所以 $\sin \alpha-\cos \alpha=\frac{\sqrt{17}}{3}$,

又因为 $\frac{1+\tan \alpha}{1-\tan \alpha}=\frac{1+\frac{\sin \alpha}{\cos \alpha}}{1-\frac{\sin \alpha}{\cos \alpha}}=\frac{\sin \alpha+\cos \alpha}{\cos \alpha-\sin \alpha}=\frac{\frac{1}{3}}{-\frac{\sqrt{17}}{3}}=-\frac{\sqrt{17}}{17}$, 故选 B.