9. (14 points) Given $\sin \alpha+\sin \beta=\frac{1}{5}, \cos \alpha+\cos \beta=\frac{1}{3}$. Find the value of $\frac{1-\cos 2(\alpha+\beta)+\sin 2(\alpha+\beta)}{1+\cos 2(\alpha+\beta)+\sin 2(\alpha+\beta)}$.


9. From $\tan \frac{\alpha+\beta}{2}=\frac{\sin \alpha+\sin \beta}{\cos \alpha+\cos \beta}=\frac{3}{5}$, we get
$$
\tan (\alpha+\beta)=\frac{2 \tan \frac{\alpha+\beta}{2}}{1-\tan ^{2} \frac{\alpha+\beta}{2}}=\frac{15}{8} .
$$

Notice that
$$
\begin{array}{l}
\tan \gamma=\frac{1-\cos 2 \gamma}{\sin 2 \gamma}=\frac{\sin 2 \gamma}{1+\cos 2 \gamma} \\
=\frac{1-\cos 2 \gamma+\sin 2 \gamma}{1+\cos 2 \gamma+\sin 2 \gamma} .
\end{array}
$$

Taking $\gamma=\alpha+\beta$, the desired result is $\frac{15}{8}$.