11. Given $\frac{\sin (\alpha+\beta)}{\sin (\alpha-\beta)}=3$. Then the value of $\frac{\tan \alpha}{\tan \beta}$ is
$$
\begin{array}{l}
\sin \alpha \cdot \cos \beta + \cos \alpha \cdot \sin \beta \\
= 3(\sin \alpha \cdot \cos \beta - \cos \alpha \cdot \sin \beta), \\
\sin \alpha \cdot \cos \beta = 2 \cos \alpha \cdot \sin \beta. \\
\text{Therefore, } \frac{\tan \alpha}{\tan \beta} = \frac{\sin \alpha \cdot \cos \beta}{\cos \alpha \cdot \sin \beta} = 2.
\end{array}
$$