Question

If $\cos \theta =\frac{\cos \alpha -e}{1-e\cos \alpha }$ then prove: $\tan \frac{\theta }{2}=\pm \sqrt{\frac{1+e}{1-e}}\tan \frac{\alpha }{2}$.

Answer 1

$\cos \theta =2{\cos }^2\frac{\theta }{2}-1$

$\Rightarrow \cos \frac{\theta }{2}=\pm \sqrt{\frac{1+\cos \theta }{2}}$

$\cos \theta =1-2{\sin }^2\theta$

$\Rightarrow \sin \frac{\theta }{2}=\pm \sqrt{\frac{1-\cos \theta }{2}}$

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

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

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

$=\pm \sqrt{\frac{(1+e)(1-\cos \alpha )}{(1-e)(1+\cos \alpha )}}$

$=\pm \sqrt{\frac{1+e}{1-e}}\cdot \sqrt{\frac{1-\cos \alpha }{1+\cos \alpha }}$

$\therefore =\pm \sqrt{\frac{1+e}{1-e}}\tan \left(\frac{\alpha }{2}\right)$