Question

If $\tan \frac{\theta }{2}=\sqrt{\left(\frac{1-e}{1+e}\right)}\tan \frac{\varphi }{2}$, prove that $\cos \varphi =\frac{\cos \theta -e}{1-e\cos \theta }$

Answer 1

We know that $\cos A=\frac{1-{\tan }^2\left(A/2\right)}{1+{\tan }^2\left(A/2\right)}$
$\therefore \cos \varphi =\frac{1-{\tan }^2\left(\varphi /2\right)}{1+{\tan }^2\left(\varphi /2\right)}=\frac{1-\left\{\right(1+e\left)/\right(1-e\left)\right\}{\tan }^2\left(\theta /2\right)}{1+\left\{\right(1+e\left)/\right(1-e\left)\right\}{\tan }^2\left(\theta /2\right)}=\frac{\{1-{\tan }^2\left(\theta /2\right)\}-e\{1+{\tan }^2\left(\theta /2\right)\}}{\{1+{\tan }^2\left(\theta /2\right)\}-e\{1-{\tan }^2\left(\theta /2\right)\}}$
from the given relation
$\therefore \cos \varphi \frac{\cos \theta -e}{1-e\cos \theta }$