9. In a right-angled triangle $ABC$, it is known that $\sin (A+B)=\frac{3}{5}, \sin (A-B)=\frac{1}{5}, AB=3$, then the area of $\triangle ABC$ is
$\qquad$ .
$$
\begin{array}{l}
\left\{\begin{array} { l } 
{ \sin ( A + B ) = \sin A \cos B + \cos A \sin B = \frac { 3 } { 5 } , } \\
{ \sin ( A - B ) = \sin A \cos B - \cos A \sin B = \frac { 1 } { 5 } }
\end{array} \Rightarrow \left\{\begin{array}{l}
\sin A \cos B=\frac{2}{5}, \\
\cos A \sin B=\frac{1}{5}
\end{array}\right.\right. \\
\Rightarrow \frac{\tan A}{\tan B}=2 \Rightarrow \tan A=2 \tan B, \text { and } \sin (A+B)=\sin C=\frac{3}{5} \Rightarrow \tan C=\frac{3}{4}, \\
\text { thus } \tan (A+B)=\frac{3 \tan B}{1-2 \tan ^{2} B}=-\frac{3}{4} \\
\Rightarrow 2 \tan ^{2} B-4 \tan B-1=0 \Rightarrow \tan B=\frac{2+\sqrt{6}}{2},
\end{array}
$$
$\tan A=2+\sqrt{6}$, as shown in the figure.
Therefore, $S_{\triangle A B C}=\frac{1}{2} \cdot 3 \cdot(2+\sqrt{6})=\frac{6+3 \sqrt{6}}{2}$.
$$