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


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.
