$If \tan \left(\frac{\theta }{2}\right)=\sqrt{\frac{1-e}{1+e}} \tan \left(\frac{\alpha }{2}\right) then \cos \alpha =$

$\begin{array}{rcl}\tan \frac{\mathrm{\theta }}{2}& =& \sqrt{\frac{1-\mathrm{e}}{1+\mathrm{e}}} \tan \frac{\mathrm{\alpha }}{2}\\ & \Rightarrow & {\tan }^2 \frac{\mathrm{\alpha }}{2} =\left(\frac{1+\mathrm{e}}{1-\mathrm{e}}\right) \tan \frac{\mathrm{\theta }}{2}\\ \cos \mathrm{\alpha }& =& \frac{1-{\tan }^2\frac{\mathrm{\alpha }}{2}}{1+{\tan }^2 {\displaystyle \frac{\mathrm{\alpha }}{2}}}\\ & =& \frac{1-\left({\displaystyle \frac{1+\mathrm{e}}{1-\mathrm{e}}}\right) {\tan }^2 {\displaystyle \frac{{\displaystyle \mathrm{\theta }}}{{\displaystyle 2}}}}{1+\left({\displaystyle \frac{1+\mathrm{e}}{1-\mathrm{e}}}\right) {\tan }^2 {\displaystyle \frac{\mathrm{\theta }}{2}}}\\ & =& \frac{1-\mathrm{e}-{\tan }^2 {\displaystyle \frac{\mathrm{\theta }}{2}}-\mathrm{e} {\tan }^2 {\displaystyle \frac{\mathrm{\theta }}{2}}}{1-\mathrm{e}+{\tan }^2 {\displaystyle \frac{\mathrm{\theta }}{2}}+\mathrm{e} {\tan }^2 {\displaystyle \frac{\mathrm{\theta }}{2}}}\end{array}$

$$\begin{gathered} =\frac{1-{\tan }^2{\displaystyle \frac{\mathrm{\theta }}{2}}-\mathrm{e}\left(1+{\tan }^2 {\displaystyle \frac{\mathrm{\theta }}{2}}\right)}{1+{\tan }^2{\displaystyle \frac{\mathrm{\theta }}{2}}-\mathrm{e}\left(1-{\tan }^2 {\displaystyle \frac{\mathrm{\theta }}{2}}\right)} \\ =\frac{{\displaystyle \frac{1-{\tan }^2 {\displaystyle \frac{\mathrm{\theta }}{2}}}{1+{\tan }^2 {\displaystyle \frac{\mathrm{\theta }}{2}}}}-\mathrm{e}}{1-\mathrm{e}\left({\displaystyle \frac{1-{\tan }^2 {\displaystyle \frac{\mathrm{\theta }}{2}}}{1+{\tan }^2 {\displaystyle \frac{\mathrm{\theta }}{2}}}}\right)} \\ =\frac{\mathrm{cos\theta }-\mathrm{e}}{1-\mathrm{e} \mathrm{cos\theta }} \end{gathered}$$