If $\alpha +\beta =\frac{\pi }{2}\text{ and }\beta +\gamma =\alpha$ then tan$\alpha$ equals

$\alpha +\beta =\frac{\pi }{2}\alpha =\frac{\pi }{2}-\beta \text{ or }\tan \alpha =\cot \beta$

or $\tan \alpha \tan \beta =1$

Again $\beta +\gamma =\alpha \Rightarrow \gamma =\alpha -\beta$

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

$\therefore \tan \alpha =\tan \beta +2\tan \gamma$