<s>[INST] $\tan \frac{\alpha}{2} \tan \frac{\beta}{2}=\frac{1-e}{1+e}$ If P is a point of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ Whose foci are S and S’.Let angle PSS’=$\alpha$ and $PS’S=\beta$ then prove that $\tan \frac{\alpha}{2} \tan \frac{\beta}{2}=\frac{1-e}{1+e}$ 
I know that S(ae,0) and S’(-ae,0) and $b^2=a^2(1-e^2)$ but don’t know how to solve further
 [/INST] Write the coefficients of $P$ as $x=a\cos t,\,y=b\sin t$ so$$\begin{align}PS^2&=a^2(\cos t-e)^2+b^2\sin^2 t\\&=a^2(1-e\cos t)^2,\\PS^{\prime2}&=a^2(1+e\cos t)^2,\\SS^{\prime2}&=4a^2e^2,\end{align}$$and$$\begin{align}\cos\alpha&=\frac{PS^2+SS^{\prime2}-PS^{\prime2}}{2PS\cdot SS^\prime}\\&=\frac{e-\cos t}{1-e\cos t},\\\cos\beta&=\frac{e+\cos t}{1+e\cos t}.\end{align}$$In terms of $A:=\tan\frac{\alpha}{2},\,B:=\tan\frac{\beta}{2}$,$$\frac{1-A^2}{1+A^2}=\frac{e-\cos t}{1-e\cos t}\implies A^2=\frac{(1-e)(1+\cos t)}{(1+e)(1-\cos t)},$$and similarly$$B^2=\frac{(1-e)(1-\cos t)}{(1+e)(1+\cos t)}$$(note we just have to replace $\cos t$ with $-\cos t$). Now take the geometric mean.
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