<s>[INST] $P$ is a point on ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ $(a>b)$ and $S$ and $S'$ are its focii If $P$ is a point on ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ $(a>b)$ and $S$ and $S'$ are its focii.
$\angle PSS'=\alpha$ and $\angle PS'S=\beta$, then prove that:
$$
\tan\left(\frac{\alpha}{2}\right)\tan\left(\frac{\beta}{2}\right)=\frac{1-e}{1+e}
$$
I am trying to do it by taking $P$ as $(0,b)$ but not able to derive required expression. Any suggestion? 
 [/INST] Another answer is, let the angle $\theta=\angle POS'$, the following identity is straightforward:
$$
\tan\left(\frac{180-\beta}{2}\right) = \sqrt{\frac{1+e}{1-e}} \tan\left(\frac{\theta}{2}\right) 
$$
and
$$
\tan\left(\frac{\alpha}{2}\right) = \sqrt{\frac{1-e}{1+e}} \tan\left(\frac{\theta}{2}\right) 
$$
Hence,
$$
\tan\left(\frac{\alpha}{2}\right)\tan\left(\frac{\beta}{2}\right)= \sqrt{\frac{1-e}{1+e}} \tan\left(\frac{\theta}{2}\right) \sqrt{\frac{1-e}{1+e}} \cot\left(\frac{\theta}{2}\right) = \frac{1-e}{1+e}
$$
</s>