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Question
If \[\tan \frac{x}{2} = \frac{\sqrt{1 - e}}{1 + e} \tan \frac{\alpha}{2}\] , then \[\cos \alpha =\]
Options
\[1 - e \cos \left( \cos x + e \right)\]
\[\frac{1 + e \cos x}{\cos x - e}\]
\[\frac{1 - e \cos x}{\cos x - e}\]
\[\frac{\cos x - e}{1 - e \cos x}\]
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Solution
\[\frac{\cos x - e}{1 - e \cos x}\]
\[\text { Given } : \tan\frac{x}{2} = \sqrt{\frac{1 - e}{1 + e}}\tan\frac{\alpha}{2}\]
\[ \Rightarrow \frac{\tan\frac{x}{2}}{\tan\frac{\alpha}{2}} = \sqrt{\frac{1 - e}{1 + e}}\]
\[\text{ Squaring both sides, we get, } \]
\[\frac{\tan^2 \frac{x}{2}}{\tan^2 \frac{\alpha}{2}} = \frac{1 - e}{1 + e}\]
\[ \Rightarrow \tan^2 \frac{\alpha}{2}\left( 1 - e \right) = \tan^2 \frac{x}{2}\left( 1 + e \right)\]
\[\Rightarrow \frac{\sin^2 \frac{\alpha}{2}}{\cos^2 \frac{\alpha}{2}}\left( 1 - e \right) = \frac{\sin^2 \frac{x}{2}}{\cos^2 \frac{x}{2}}\left( 1 + e \right)\]
\[ \Rightarrow \frac{\frac{1}{2}\left( 1 - cos\alpha \right)}{\frac{1}{2}\left( 1 + cos\alpha \right)}\left( 1 - e \right) = \frac{\frac{1}{2}\left( 1 - \text{ cos } x \right)}{\frac{1}{2}\left( 1 + \text{ cos } x \right)}\left( 1 + e \right)\]
\[ \Rightarrow \left( 1 - cos\alpha \right)\left( 1 + \text{ cos } x \right)\left( 1 - e \right) = \left( 1 + cos\alpha \right)\left( 1 - \text{ cos } x \right)\left( 1 + e \right)\]
\[ \Rightarrow \left( 1 + \text{ cos } x \right)\left( 1 - e \right) - cos\alpha\left( 1 + \text{ cos } x \right)\left( 1 - e \right) = \left( 1 - \text{ cos } x \right)\left( 1 + e \right) + cos\alpha\left( 1 - \text{ cos } x \right)\left( 1 + e \right)\]
\[ \Rightarrow cos\alpha\left\{ \left( 1 + \text{ cos } x \right)\left( 1 - e \right) + \left( 1 - \text{ cos } x \right)\left( 1 + e \right) \right\} = \left( 1 + \text{ cos } x \right)\left( 1 - e \right) - \left( 1 - \text{ cos } x \right)\left( 1 + e \right)\]
\[ \Rightarrow cos\alpha = \frac{2\text{ cos } x - 2e}{2 - 2ecosx} = \frac{\text{ cos } x - e}{1 - ecosx}\]
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