Question

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

Answer 1

$L.H.S={\tan }^{-1}\frac{2\tan \frac{\alpha }{2}\tan \left(\frac{pi}{4}\frac{\beta }{2}\right)}{1-{\tan }^2\frac{\alpha }{2}{\tan }^2\left(\frac{pi}{4}\frac{\beta }{2}\right)}(since2{\tan }^{-1}x={\tan }^{-1}\frac{2x}{1-x^2})$
$={\tan }^{-1}\frac{2\tan \frac{\alpha }{2}\frac{1-\tan \frac{\beta }{2}}{1+\tan \frac{\beta }{2}}}{1-{\tan }^2\frac{\alpha }{2}{\left(\frac{1-\tan \frac{\beta }{2}}{1+\tan \frac{\beta }{2}}\right)}^2}$
$={\tan }^{-1}\frac{2\tan \frac{\alpha }{2}(1-\tan \frac{\beta }{2})}{{(1+\tan \frac{\beta }{2})}^2-{\tan }^2\frac{\alpha }{2}{(1-\tan \frac{\beta }{2})}^2}$
$={\tan }^{-1}\frac{2\tan \frac{\alpha }{2}(1-\tan \frac{\beta }{2})}{{(1+\tan \frac{\beta }{2})}^{}(1-\tan \frac{\alpha }{2})+2{\tan }^2\frac{\beta }{2}{(1+\tan \frac{\alpha }{2})}^{}}$
$={\tan }^{-1}\frac{\frac{2\tan \frac{\alpha }{2}1-{\tan }^2\frac{\beta }{2}}{1+{\tan }^2\frac{\alpha }{2}1+{\tan }^2\frac{\beta }{2}}}{\frac{1-{\tan }^2\frac{\alpha }{2}}{1+{\tan }^2\frac{\alpha }{2}}\frac{2\tan \frac{\beta }{2}}{1+{\tan }^2\frac{\beta }{2}}}$
$={\tan }^{-1}\left(\frac{\sin \alpha \cos \beta }{\cos \alpha +\sin \beta }\right)=R.H.S$