$P$ 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? 
 A: Using Analytic Geometry (high school): Why is the sum of the distances from any point of the ellipse to the two foci the major axis?, $S'P+SP=2a$
Again using The Law of Sines (or Sine Rule),
$$\dfrac{SP}{\sin\alpha}=\dfrac{S'P}{\sin\beta}=\dfrac{S'S}{\sin\{\pi-(\alpha+\beta)\}}=\dfrac{2ae}{\sin(\alpha+\beta)}$$
Again $$\dfrac{SP}{\sin\alpha}=\dfrac{S'P}{\sin\beta}=\dfrac{S'P+SP}{\sin\alpha+\sin\beta}=\dfrac{2a}{\sin\alpha+\sin\beta}$$
$$\implies\dfrac{2a}{\sin\alpha+\sin\beta}=\dfrac{2ae}{\sin(\alpha+\beta)}$$
$$\iff e=\dfrac{\sin(\alpha+\beta)}{\sin\alpha+\sin\beta}=\dfrac{2\sin\dfrac{\alpha+\beta}2\cos\dfrac{\alpha+\beta}2}{2\sin\dfrac{\alpha+\beta}2\cos\dfrac{\alpha-\beta}2}=\dfrac{\cos\dfrac{\alpha+\beta}2}{\cos\dfrac{\alpha-\beta}2}$$
Now use Componendo and Dividendo
A: Hint:
$$
PS = \frac{a(1-e^2)}{1-e \cos(\alpha)}
$$
$$
PS' = \frac{a(1-e^2)}{1-e \cos(\beta)}
$$
$$
\frac{PS}{\sin(\beta)} = \frac{PS'}{\sin(\alpha)}
$$
A: 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,
$$
\tan\left(\frac{\alpha}{2}\right)\tan\left(\frac{\beta}{2}\right)= \sqrt{\frac{1-e}{1+e}} \tan\left(\frac{\theta}{2}\right) \sqrt{\frac{1-e}{1+e}} \cot\left(\frac{\theta}{2}\right) = \frac{1-e}{1+e}
$$
