9. Given $A, B \in$
$\left(0, \frac{\pi}{2}\right)$, and $\frac{\sin A}{\sin B}=\sin (A+B)$, find the maximum value of $\tan A$.
$$
\begin{array}{l}
\frac{\sin A}{\sin B}=\sin A \cos B+\cos A \sin B \Rightarrow \sin A=\sin A \sin B \cos B+\cos A \sin ^{2} B \\
\Rightarrow \tan A=\tan A \sin B \cos B+\sin ^{2} B \Rightarrow \tan A=\frac{\sin ^{2} B}{1-\sin B \cos B} \\
=\frac{\sin ^{2} B}{\sin ^{2} B+\cos ^{2} B-\sin B \cos B}=\frac{\tan ^{2} B}{\tan ^{2} B-\tan B+1}=\frac{1}{\frac{1}{\tan ^{2} B}-\frac{1}{\tan B}+1}
\end{array}
$$
$$
=\frac{1}{\left(\frac{1}{\tan B}-\frac{1}{2}\right)^{2}+\frac{3}{4}} \leqslant \frac{4}{3} \text {, equality holds when } \tan B=2 \text {. Therefore, the maximum value of } \tan A \text { is } \frac{4}{3} \text {. }
$$