If tan β = tan α + tan γ 1 + tan α tan γ then sin 2 α + sin 2 γ 1 + sin 2…

Given, $\tan \beta =\frac{\tan \alpha +\tan \gamma }{1+\tan \alpha \tan \gamma }=\frac{{\displaystyle \frac{\sin \alpha }{\cos \alpha }}+{\displaystyle \frac{\sin \gamma }{\cos \gamma }}}{1+{\displaystyle \frac{\sin \alpha }{\cos \alpha }}.{\displaystyle \frac{\sin \gamma }{\cos \gamma }}}$

$=\frac{\sin \alpha .\cos \gamma +\sin \gamma .\cos \alpha }{\cos \alpha .\cos \gamma +\sin \alpha .\sin \gamma }$

$\Rightarrow \tan \beta =\frac{\sin \left(\alpha +\gamma \right)}{\cos \left(\alpha -\gamma \right)}\dots \left(1\right)$

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

Substituting value of $\tan \beta$ from equation $\left(1\right)$,

$\sin 2\beta =\frac{2\times {\displaystyle \frac{\sin \left(\alpha +\gamma \right)}{\cos \left(\alpha -\gamma \right)}}}{1+{\left({\displaystyle \frac{\sin \left(\alpha +\gamma \right)}{\cos \left(\alpha -\gamma \right)}}\right)}^2}$

$=2\left[\frac{\sin \left(\alpha +\gamma \right).\cos \left(\alpha -\gamma \right)}{{\cos }^2\left(\alpha -\gamma \right)+{\sin }^2\left(\alpha +\gamma \right)}\right]$

                               $=2\left[\frac{\left(\sin \alpha .\cos \gamma +\cos \alpha .\sin \gamma \right)\left(\cos \alpha .\cos \gamma +\sin \alpha .\sin \gamma \right)}{{\left(\cos \alpha .\cos \gamma +\sin \alpha .\sin \gamma \right)}^2+{\left(\sin \alpha .\cos \gamma +\cos \alpha .\sin \gamma \right)}^2}\right]$

$=2\left[\frac{\sin \alpha .\cos \alpha .{\cos }^2\gamma +{\sin }^2 \alpha .\cos \gamma .\sin \gamma +{\cos }^2 \alpha .\cos \gamma .\sin \gamma +{\sin }^2\gamma .\cos \alpha .\sin \alpha }{{\cos }^2\gamma +{\sin }^2\gamma +4\sin \alpha .\cos \alpha .\sin \gamma .\cos \gamma }\right]$

$=2\left[\frac{\sin \alpha .\cos \alpha +\sin \gamma .\cos \gamma }{1+\sin 2\alpha .\sin 2\gamma }\right]$

$=\frac{\sin 2\alpha +\sin 2\gamma }{1+\sin 2\alpha .\sin 2\gamma }$  Proved.