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Question
If \[2 \tan\frac{\alpha}{2} = \tan\frac{\beta}{2}\] , prove that \[\cos \alpha = \frac{3 + 5 \cos \beta}{5 + 3 \cos \beta}\]
Solution
\[RHS = \frac{3 + 5\cos \beta}{5 + 3\cos \beta}\]
\[ = \frac{3 + 5\left( \frac{1 - \tan^2 \frac{\beta}{2}}{1 + \tan^2 \frac{\beta}{2}} \right)}{5 + 3\left( \frac{1 - \tan^2 \frac{\beta}{2}}{1 + \tan^2 \frac{\beta}{2}} \right)}\]
\[ = \frac{3 + 3 \tan^2 \frac{\beta}{2} + 5 - 5 \tan^2 \frac{\beta}{2}}{5 + 5 \tan^2 \frac{\beta}{2} + 3 - 3\tan \frac{\beta}{2}}\]
\[ = \frac{8 - 2 \tan^2 \frac{\beta}{2}}{8 + 2 \tan^2 \frac{\beta}{2}}\]
\[ = \frac{8 - 8 \tan^2 \frac{\alpha}{2}}{8 + 8 \tan^2 \frac{\alpha}{2}} \left[ \because 2\tan \frac{\alpha}{2} = \tan \frac{\beta}{2} \right]\]
\[ = \frac{8\left( 1 - \tan^2 \frac{\alpha}{2} \right)}{8\left( 1 + \tan^2 \frac{\alpha}{2} \right)}\]
\[ = \frac{1 - \tan^2 \frac{\alpha}{2}}{1 + \tan^2 \frac{\alpha}{2}}\]
\[ = \cos \alpha = LHS\]
\[\text{ Hence proved } .\]
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