Question

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

Answer 1

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

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

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

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

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

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

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

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

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

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

LHS=RHS

Hence proof