Question

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

• A. True
• B. False

Answer 1

The correct option is A True

$L.H.S={\tan }^{-1}\frac{2\tan \frac{\alpha }{2}\tan (\frac{\pi }{4}-\frac{\beta }{2})}{1-{\tan }^2\frac{\alpha }{2}{\tan }^2(\frac{\pi }{4}-\frac{\beta }{2})}$ since $2{\tan }^{-1}x={\tan }^{-1}\frac{2x}{1-x^2}$

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

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

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

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

$={\tan }^{-1}\left(\frac{\sin \alpha \cos \beta }{\cos \alpha +\sin \beta }\right)=R.H.S$

Hence proved.

Hence the given statement is true.