Question

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

• A. $\frac{1-e\cos \theta }{\cos \theta +e}$
• 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 $\frac{\cos \theta -e}{1-e\cos \theta }$

Simplifying the above equation, we get

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

Therefore

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

Therefore

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

$\frac{1-e+\cos \theta -e\cos \theta -(1+e-\cos \theta -e\cos \theta )}{1-e+\cos \theta -e\cos \theta +1+e-\cos \theta -e\cos \theta }=\cos \alpha$ $\left[\text{using componendo and dividendo}\right]$

$\frac{2(\cos \theta -e)}{2(1-e\cos \theta )}=\cos \alpha$
Therefore

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