Question

If $2\tan \frac{\alpha }{2}=\tan \frac{\beta }{2}$, prove that $\cos \alpha =\frac{3+5\cos \beta }{5+3\cos \beta }$

Answer 1

$$\begin{gathered} \mathrm{RHS}=\frac{3+5\cos \beta }{5+3\cos \beta } \\ =\frac{3+5\left({\displaystyle \frac{1-{\tan }^2{\displaystyle \frac{\beta }{2}}}{1+{\tan }^2{\displaystyle \frac{\beta }{2}}}}\right)}{5+3\left({\displaystyle \frac{1-{\tan }^2{\displaystyle \frac{\beta }{2}}}{1+{\tan }^2{\displaystyle \frac{\beta }{2}}}}\right)} \\ =\frac{3+3{\tan }^2{\displaystyle \frac{\beta }{2}}+5-5{\tan }^2{\displaystyle \frac{\beta }{2}}}{5+5{\tan }^2{\displaystyle \frac{\beta }{2}}+3-3\tan {\displaystyle \frac{\beta }{2}}} \\ =\frac{8-2{\tan }^2{\displaystyle \frac{\beta }{2}}}{8+2{\tan }^2{\displaystyle \frac{\beta }{2}}} \\ =\frac{8-8{\tan }^2{\displaystyle \frac{\alpha }{2}}}{8+8{\tan }^2{\displaystyle \frac{\alpha }{2}}}\left[\because 2\tan \frac{\alpha }{2}=\tan \frac{\beta }{2}\right] \\ =\frac{8\left(1-{\tan }^2{\displaystyle \frac{\alpha }{2}}\right){\displaystyle }}{8\left(1+{\tan }^2{\displaystyle \frac{\alpha }{2}}\right)} \\ =\frac{1-{\tan }^2{\displaystyle \frac{\alpha }{2}}{\displaystyle }}{1+{\tan }^2{\displaystyle \frac{\alpha }{2}}} \\ =\cos \alpha =\mathrm{LHS} \\ \mathrm{Hence}\mathrm{proved}. \end{gathered}$$