Question

prove that $2{\tan }^{-1}\left[\tan \frac{\alpha }{2}\tan \left(\frac{\pi }{4}-\frac{\beta }{2}\right)\right]={\tan }^{-1}\frac{\sin \alpha cos\beta }{\cos \alpha +\sin \beta }$

Answer 1

$2{\tan }^{-1}\left[\tan \frac{\alpha }{2}\tan \frac{\pi }{4}-\frac{\beta }{2}\right]$

$=2{\tan }^{-1}\left[\tan \frac{\alpha }{2}\right(\frac{\tan \frac{\pi }{4}-\tan \frac{\beta }{2}}{1+\tan \frac{\pi }{4}\tan \frac{\beta }{2}}\left)\right]$

$=2{\tan }^{-1}\left[\tan \frac{\alpha }{2}\right(\frac{1-\tan \frac{\beta }{2}}{1+\tan \frac{\beta }{2}}\left)\right]$

$={\tan }^{-1}\frac{\frac{2\tan \frac{\alpha }{2}(1-\tan \frac{\beta }{2})}{1+\tan \frac{\beta }{2}}}{1-{\left(\frac{2\tan \frac{\alpha }{2}(1-\tan \frac{\beta }{2})}{1+\tan \frac{\beta }{2}}\right)}^2}$

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

Convert the $\tan \frac{\alpha }{2}=\frac{\sin \frac{\alpha }{2}}{\cos \frac{\alpha }{2}}\&\tan \frac{\beta }{2}=\frac{\sin \frac{\beta }{2}}{\cos \frac{\beta }{2}}$ you will get the desired value.

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