Question

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

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

Answer 1

The correct option is D

$\frac{cos\theta -e}{1-ecos\theta }$

Given: $tan\frac{\theta }{2}=\sqrt{\frac{1-e}{1+e}}.tan\frac{\alpha }{2}\Rightarrow \frac{tan\frac{\theta }{2}}{tan\frac{\alpha }{2}}=\sqrt{\frac{1-e}{1+e}}$

Squaring both sides, we get,

$\frac{ta{n}^2\frac{\theta }{2}}{ta{n}^2\frac{\alpha }{2}}=\frac{1-e}{1+e}\Rightarrow ta{n}^2\frac{\alpha }{2}(1-e)=ta{n}^2\frac{\theta }{2}(1+e)\Rightarrow \frac{si{n}^2\frac{\alpha }{2}}{co{s}^2\frac{\alpha }{2}}(1-e)=\frac{si{n}^2\frac{\theta }{2}}{co{s}^2\frac{\theta }{2}}(1+e)\Rightarrow \frac{\frac{1}{2}(1-cos\alpha )}{\frac{1}{2}(1+cos\alpha )}(1-e)$

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

$\Rightarrow cos\alpha =\frac{2cos\theta -2e}{2-2cos\theta }=\frac{cos\theta -e}{1-ecos\theta }$