If the normals at the points of vectorial angles $\alpha,\beta,\gamma,\delta$ on
the conic $l/r=1+e\cos\theta$
meet at a point, then prove that
$\tan(\alpha/2)\tan(\beta/2)\tan(\gamma/2)\tan(\delta/2)+(\frac{1+e}{1-e})^2 =0$
Also prove that if the normals meet at a point $(\rho,\phi)$, then
$\alpha+\beta+\gamma+\delta-2\phi=$ odd multiple of $\pi$.
MY ATTEMPT
Equation of normal to the given conic at a point whose vectorial angle is $\alpha$ and $\beta$ are $$\frac{le \sin\alpha}{1+e \cos\alpha}r = e\sin\theta+\sin(\theta-\alpha)$$ And $$\frac{le \sin\beta}{1+e \cos\beta}r = e\sin\theta+\sin(\theta-\beta) $$respectively
Subtracting these equation and solving We get $$r\cos(\theta-\frac{\alpha+\beta}{2})=\frac{l(1-e^2)\cos\frac{(\beta-\alpha)}{2}}{(1–e\cos\alpha)](1-e\cos\beta)}$$
How to obtain $\tan\frac{\alpha}{2}\tan\frac{\beta}{2} = \frac{1+e}{1–e}$from above results
Am i in the right path?