11 Given $\frac{\sin (\alpha+\beta)}{\sin (\alpha-\beta)}=3$, then the value of $\frac{\tan \alpha}{\tan \beta}$ is
11 From the given, we have
$$
\sin \alpha \cos \beta + \cos \alpha \sin \beta = 3(\sin \alpha \cos \beta - \cos \alpha \sin \beta),
$$

which means
$$
\sin \alpha \cos \beta = 2 \cos \alpha \sin \beta,
$$

so
$$
\frac{\tan \alpha}{\tan \beta} = \frac{\sin \alpha \cos \beta}{\cos \alpha \sin \beta} = 2.
$$