9. In $\triangle P F_{1} F_{2}$, using the Law of Sines and the Ratio Theorem, we have $\frac{\left|P F_{1}\right|}{\sin \beta}=\frac{\left|P F_{2}\right|}{\sin \alpha}=\frac{\left|F_{1} F_{2}\right|}{\sin (\alpha+\beta)}=$
$$
\begin{array}{l}
\frac{\left|P F_{1}\right|-\left|P F_{2}\right|}{\sin \beta-\sin \alpha} \text {, i.e., } \frac{2 a}{2 \cos \frac{\beta+\alpha}{2} \sin \frac{\beta-\alpha}{2}}=\frac{2 c}{2 \sin \frac{\alpha+\beta}{2} \cos \frac{\alpha+\beta}{2}}, \frac{\sin \frac{\beta-\alpha}{2}}{\sin \frac{\beta+\alpha}{2}}=\frac{a}{c}, \\
\frac{\sin \frac{\beta}{2} \cos \frac{\alpha}{2}-\cos \frac{\beta}{2} \sin \frac{\alpha}{2}}{\sin \frac{\beta}{2} \cos \frac{\alpha}{2}+\cos \frac{\beta}{2} \sin \frac{\alpha}{2}}=\frac{a}{c}, \frac{\tan \frac{\beta}{2}-\tan \frac{\alpha}{2}}{\tan \frac{\beta}{2}+\tan \frac{\alpha}{2}}=\frac{a}{c}, \\
\text { By the Componendo and Dividendo Theorem, } \frac{2 \tan \frac{\beta}{2}}{2 \tan \frac{\alpha}{2}}=\frac{a+c}{c-a}, \frac{\tan \frac{\alpha}{2}}{\tan \frac{\beta}{2}}=\frac{c-a}{c+a} .
\end{array}
$$
9. Point $P$ is any point on the right branch of the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$, $F_{1}, F_{2}$ are the left and right foci, respectively. Let $\angle P F_{1} F_{2}=$ $\alpha, \angle P F_{2} F_{1}=\beta$, try to express $\frac{\tan \frac{\alpha}{2}}{\tan \frac{\beta}{2}}$ in terms of $a$ and $b$.