16. 已知 $\alpha 、 \beta \in(0, \pi)$ 且 $\tan (\alpha-\beta)=\frac{1}{2}, \cos \beta=-\frac{7 \sqrt{2}}{10}$, 则 $\tan (2 \alpha-\beta)=$
$\because 0<\beta<\pi ， \cos \beta=-\frac{7 \sqrt{2}}{10}$ ，

$\therefore \sin \beta=\sqrt{1-\cos ^{2} \beta}=\frac{\sqrt{2}}{10}$,

$\therefore \tan \beta=-\frac{1}{7}$.

又 $\because \tan (\alpha-\beta)=\frac{1}{2}$ ，

$\therefore \tan \alpha=\tan [(\alpha-\beta)+\beta]=\frac{\tan (\alpha-\beta)+\tan \beta}{1-\tan (\alpha-\beta) \tan \beta}$

$=\frac{\frac{1}{2}+\left(-\frac{1}{7}\right)}{1-\frac{1}{2} \times\left(-\frac{1}{7}\right)}=\frac{1}{3}$.

又 $\tan (2 \alpha-\beta)=\tan [(\alpha-\beta)+\alpha]=\frac{\tan (\alpha-\beta)+\tan \alpha}{1-\tan (\alpha-\beta) \tan \alpha}$

$=\frac{\frac{1}{2}+\frac{1}{3}}{1-\frac{1}{2} \times \frac{1}{3}}=1$.

## 答案: 1