Eccentricity of ellipse in terms of eccentric angles of extremities of focal chord?

Question

If $A$ and $B$ are eccentric angles of the extremities of a focal chord of an ellipse then eccentricity of the ellipse is?

The answer given is $\dfrac{\sin A + \sin B}{\sin (A+B)}$.

Now I tried to do this by assuming extremities to be $P(a\cos A, b\sin A)$ and $Q (a\cos B, b\sin B)$ and focus to be $F(ae , 0)$. By equating slopes of $PF$ and $FQ$, I got $e=\dfrac{\sin (A+B)}{\sin A -\sin B}$. I can't find any way to get the required answer from this. Was my approach right? What should I do?

Answer 1

The equation of chord joining pairs of point $[A]$ and $[B]$ is:

$$\frac{x}{a} \cos \frac{A+B}{2}+\frac{y}{b} \sin \frac{A+B}{2} =\cos \frac{A-B}{2}$$

Put $(x,y)=(\pm ae,0)$, then

$$e=\pm \frac{\cos \frac{A-B}{2}}{\cos \frac{A+B}{2}}$$

Also,

\begin{align} \frac{\sin A+\sin B}{\sin (A+B)} &= \frac{2\sin \frac{A+B}{2} \cos \frac{A-B}{2}} {2\sin \frac{A+B}{2} \cos \frac{A+B}{2}} \\ &= \frac{\cos \frac{A-B}{2}}{\cos \frac{A+B}{2}} \\ \frac{\sin (A-B)}{\sin A-\sin B} &= \frac{2\sin \frac{A-B}{2} \cos \frac{A-B}{2}} {2\sin \frac{A-B}{2} \cos \frac{A+B}{2}} \\ &= \frac{\cos \frac{A-B}{2}}{\cos \frac{A+B}{2}} \\ \end{align}

In conclusion: $$\fbox{ $e =\pm \frac{\cos \frac{A-B}{2}}{\cos \frac{A+B}{2}} =\pm \frac{\sin A+\sin B}{\sin (A+B)} =\pm \frac{\sin (A-B)}{\sin A-\sin B} =\pm \frac{1+\tan \frac{A}{2} \tan \frac{B}{2}} {1-\tan \frac{A}{2} \tan \frac{B}{2}} \, $}$$

See also another forum here.