If $\mathrm{\alpha }, \mathrm{\beta }$ are the ends of a focal chord of the $\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}+\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1,$ then its eccentricity e is

$$\begin{gathered} \mathrm{If} \mathrm{\alpha } \mathrm{and} \mathrm{\beta } \mathrm{are} \mathrm{ends} \mathrm{of} \mathrm{focal} \mathrm{chord} \mathrm{of} \mathrm{ellipse} \mathrm{then} \\ \mathrm{e}=\frac{\cos \left({\displaystyle \frac{\mathrm{\alpha }-\mathrm{\beta }}{2}}\right)}{\cos \left(\frac{\mathrm{\alpha }+\mathrm{\beta }}{2}\right)} \\ \mathrm{e}=\frac{\cos {\displaystyle \frac{\mathrm{\alpha }}{2}} \cos {\displaystyle \frac{\mathrm{\beta }}{2}}+\sin {\displaystyle \frac{\mathrm{\alpha }}{2}} \sin {\displaystyle \frac{\beta }{2}}}{\cos \frac{\mathrm{\alpha }}{2} \cos \frac{\beta }{2}-\sin \frac{\mathrm{\alpha }}{2} \sin \frac{\beta }{2}} \\ \mathrm{e}=\frac{1+\mathrm{Tan}\frac{\mathrm{\alpha }}{2} \mathrm{Tan}\raisebox{1ex}{$ \mathrm{\beta }$}\!\left/ \!\raisebox{-1ex}{$2$}\right. }{1-\mathrm{Tan}\frac{\mathrm{\alpha }}{2} \mathrm{Tan}\raisebox{1ex}{$ \mathrm{\beta }$}\!\left/ \!\raisebox{-1ex}{$2$}\right. } \end{gathered}$$