If the eccentric angles of the extremities of a focal chord (other than the major axis) of the ellipse $\frac{{\mathrm{x}}^2}{25}+\frac{{\mathrm{y}}^2}{9}=1$ are $\mathrm{\alpha }$ and $\mathrm{\beta }$ then $\frac{\mathrm{Cot}(\mathrm{\alpha }/2)}{\mathrm{Tan}(\mathrm{\beta }/2)}=\dots \dots .$

equation of  of chord joining $\alpha and \beta is$

$$\begin{gathered} \frac{\mathrm{x}}{5}\cos \left(\frac{\mathrm{\alpha }+\mathrm{\beta }}{2}\right)+\frac{\mathrm{y}}{3}\sin \left(\frac{\mathrm{\alpha }+\mathrm{\beta }}{2}\right) \\ =\cos \left(\frac{\mathrm{\alpha }-\mathrm{\beta }}{2}\right) \\ \mathrm{passes} \mathrm{through} \mathrm{focus} (\mathrm{ae}, 0) \\ =(\sqrt{25-9}, 0) \\ =(4, 0) \\ \frac{4}{5}\cos \left(\frac{\mathrm{\alpha }+\mathrm{\beta }}{2}\right)=\cos \left(\frac{\mathrm{\alpha }-\mathrm{\beta }}{2}\right) \\ \frac{4}{5}=\frac{\cos \left(\frac{\mathrm{\alpha }-\mathrm{\beta }}{2}\right)}{\cos \left(\frac{\mathrm{\alpha }+\mathrm{\beta }}{2}\right)} \\ By \mathrm{apply}ing \mathrm{comp}onendo \& \mathrm{div}i\mathrm{dendo} rule we get \\ \frac{9}{-1}=\frac{2\cos {\displaystyle \frac{\mathrm{\alpha }}{2}} \cos {\displaystyle \raisebox{1ex}{$\mathrm{\beta }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{2\sin \frac{\mathrm{\alpha }}{2} \sin \raisebox{1ex}{$\mathrm{\beta }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.} \\ \frac{\cot {\displaystyle \raisebox{1ex}{$\mathrm{\alpha }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{\mathrm{tab}{\displaystyle \raisebox{1ex}{$ \mathrm{\beta }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}=-9 \end{gathered}$$