Question

If $\tan \frac{\theta }{2}=\sqrt{\frac{1-e}{1+e}}\tan \frac{\alpha }{2},$ then $\cos \alpha$ =

• A. $1-e\cos (cos\theta +e)$
• B. $\frac{1+e\cos \theta }{\cos \theta -e}$
• C. $\frac{1-e\cos \theta }{\cos \theta -e}$
• D. $\frac{\cos \theta -e}{1-e\cos \theta }$

Answer 1

The correct option is D $\frac{\cos \theta -e}{1-e\cos \theta }$

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

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

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

$\Rightarrow \left(\frac{1-\cos \theta }{1+\cos \theta }\right)\times \left(\frac{1+e}{1-e}\right)=\left(\frac{1-\cos \alpha }{1+\cos \alpha }\right)$

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

apply componendo and dividendo,

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

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

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

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

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