ellipse angles at foci formula

Question

I read (in multiples sources) that for any point P on an ellipse $\frac{\tan\frac{\phi}{2}}{\tan\frac{\theta}{2}}=\frac{1-e}{1+e}$ where e is the eccentricity of the ellipse and $\theta$ and $\phi$ are as in the image below. How is this formula derived from plane trigonometric formulas? I tried using the law of tangents or the law of sines on the triangle $F_1F_2P$ together with the fact that $PF_2=\frac{a(1-e^2)}{1+e\cos\theta}$ but no luck. .

Answer 1

Start from $$ PF_1=\frac{a(1-e^2)}{1-e\cos\phi},\quad PF_2=\frac{a(1-e^2)}{1+e\cos\theta} $$ and use $PF_1+PF_2=2a$ to obtain $$ \frac{1}{1-e\cos\phi}+\frac{1}{1+e\cos\theta}={2\over1-e^2}. $$ Now substitute the identities: $$ \cos\phi={1-\tan^2{\phi\over2}\over1+\tan^2{\phi\over2}}, \quad \cos\theta={1-\tan^2{\theta\over2}\over1+\tan^2{\theta\over2}} $$ and expand everything: you'll get your formula.