Question

If $\cos \alpha =\frac{2\cos \beta -1}{2-\cos \beta }$, $(0<\alpha ,\beta ,\pi )$ then $\frac{\tan \left(\alpha /2\right)}{\tan \left(\beta /2\right)}$=

Answer 1

$\cos \alpha =\frac{2\cos \beta -1}{2-\cos \beta }$

$\frac{1-{\tan }^2\frac{\alpha }{2}}{1+{\tan }^2\frac{\alpha }{2}}=\frac{2-2{\tan }^2\frac{\beta }{2}-1-{\tan }^2\frac{\beta }{2}}{2+2{\tan }^2\frac{\beta }{2}-1+{\tan }^2\frac{\beta }{2}}$

$\frac{1-{\tan }^2\frac{\alpha }{2}}{1+{\tan }^2\frac{\alpha }{2}}=\frac{1-3{\tan }^2\frac{\beta }{2}}{1+3{\tan }^2\frac{\beta }{2}}$

$\frac{{\tan }^2\frac{\alpha }{2}}{\tan \frac{\beta }{2}}=1$.