Question

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

Answer 1

$\cos \theta =\frac{\cos \alpha -e}{1-e\cos \alpha }$

$\Rightarrow \cos \theta (1-e\cos \alpha )=\cos \alpha -e$

$\Rightarrow \cos \theta -e\cos \theta \cos \alpha =\cos \alpha -e$

$\Rightarrow e-e\cos \theta \cos \alpha =\cos \alpha -\cos \theta$

$\Rightarrow e(\cos \theta \cos \alpha -1)=\cos \theta -\cos \alpha$

$\frac{1}{e}=\frac{\cos \theta \cos \alpha -1}{\cos \theta -\cos \alpha }$

$\frac{1+e}{1-e}=\frac{\cos \theta \cos \alpha -1+\cos \theta -\cos \alpha }{\cos \theta \cos \alpha -1-\cos \theta +\cos \alpha }$

$=\frac{\cos \theta (1+\cos \alpha )-(1+\cos \alpha )}{\cos \alpha (1+\cos \theta )-(1+\cos \theta )}$

$=\frac{(\cos \theta -1)(1+\cos \alpha )}{(\cos \alpha -1)(1+\cos \theta )}$

$=\frac{1+\cos \alpha }{1-\cos \alpha }$$\frac{1-\cos \theta }{1+\cos \theta }$

$[{\tan }^{2}A=\frac{2{\sin }^{2}A}{2{\cos }^{2}A}]$

${\tan }^{2}A=\frac{1-1+2{\sin }^{2}A}{1+2co{s}^{2}A-1}=\frac{1-(1-2{\sin }^{2}A)}{1+(2{\cos }^{2}A-1)}$

$=\frac{1-\cos 2A}{1+\cos 2A}$

$=\frac{1+e}{1-e}=\frac{{\tan }^2\left(\frac{\theta }{2}\right)}{{\tan }^2\left(\frac{\alpha }{2}\right)}$

$=\pm \sqrt{\frac{1+e}{1-e}}=\frac{\tan \frac{\theta }{2}}{\tan \alpha 2}$

$\Rightarrow \pm (\tan \frac{\alpha }{2}\times \sqrt{\frac{1+e}{1-e}})=\tan \frac{\theta }{2}$

$\Rightarrow (\tan \frac{\theta }{2}\pm \sqrt{\frac{1+e}{1-e}}\tan \frac{\alpha }{2})=0$

$\Rightarrow (\tan \frac{\theta }{2}+\sqrt{\frac{1+e}{1-e}}\tan \frac{\alpha }{2})(\tan \frac{\theta }{2}-\sqrt{\frac{1+e}{1-e}}\tan \frac{\alpha }{2})=0$