Example 1 Let $P$ be a point on the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$, $F_{1}, F_{2}$ be the two foci of the ellipse, and $e$ be the eccentricity of the ellipse. If $\angle P F_{1} F_{2}=\alpha, \angle P F_{2} F_{1}=\beta$, try to express $\tan \frac{\alpha}{2} \cdot \tan \frac{\beta}{2}$ in terms of $e$.
Solve As shown in Figure 8-1, in $\triangle P F_{1} F_{2}$, by the Law of Sines:
$$
\frac{\left|P F_{1}\right|}{\sin \beta}=\frac{\left|P F_{2}\right|}{\sin \alpha}=\frac{\left|F_{1} F_{2}\right|}{\sin (\alpha+\beta)},
$$

By the property of the sum of ratios, we have $\frac{\left|P F_{1}\right|+\left|P F_{2}\right|}{\sin \alpha+\sin \beta}=\frac{\left|F_{1} F_{2}\right|}{\sin (\alpha+\beta)}$, which means $\frac{\sin (\alpha+\beta)}{\sin \alpha+\sin \beta}=\frac{2 c}{2 a}=e, \frac{2 \sin \frac{\alpha+\beta}{2} \cos \frac{\alpha+\beta}{2}}{2 \sin \frac{\alpha+\beta}{2} \cos \frac{\alpha-\beta}{2}}=e, \frac{\cos \frac{\alpha}{2} \cos \frac{\beta}{2}-\sin \frac{\alpha}{2} \sin \frac{\beta}{2}}{\cos \frac{\alpha}{2} \cos \frac{\beta}{2}+\sin \frac{\alpha}{2} \sin \frac{\beta}{2}}=e$, thus $\tan \frac{\alpha}{2} \cdot \tan \frac{\beta}{2}=\frac{1-e}{1+e}$.