9. Given $\angle A, \angle B \in\left(0, \frac{\pi}{2}\right)$, and $\frac{\sin A}{\sin B}=$ $\sin (A+B)$. Find the maximum value of $\tan A$.
$$
\begin{array}{l}
\sin A=\sin B(\sin A \cdot \cos B+\cos A \cdot \sin B) \\
\Rightarrow \tan A=\sin B(\tan A \cdot \cos B+\sin B) . \\
\text { Let } t=\tan A \text {. Then } \\
t=t \sin B \cdot \cos B+\sin ^{2} B \\
\Rightarrow 2 t=t \sin 2 B+1-\cos 2 B \\
\Rightarrow 2 t-1=\sqrt{t^{2}+1} \sin (2 B-\theta),
\end{array}
$$

where $\theta$ is an acute angle, $\sin \theta=\frac{1}{\sqrt{t^{2}+1}}$.
Therefore, $|2 t-1| \leqslant \sqrt{t^{2}+1}$.
Noting that $t>0$, we get $t \leqslant \frac{4}{3}$.
When $\angle A=\arctan \frac{4}{3}, \angle B=\frac{1}{2}\left(\frac{\pi}{2}+\arcsin \frac{3}{5}\right)$, the given equation holds.
Thus, the maximum value of $\tan A$ is $\frac{4}{3}$.