If tan $$\beta \, = \, \dfrac{\tan \, \alpha \, + \, \tan \, \gamma}{1 \, + \, \tan \, \alpha \,\tan \, \gamma},$$ Prove that:  $$\sin \, 2\beta \, = \, \dfrac{\sin \, 2\alpha \, + \, \sin \, 2\gamma}{1 \, + \, \sin \, 2\alpha \, \sin \, 2\gamma}$$

Solution

$$tan \, \beta  \, = \, \dfrac{sin \, (\alpha  \, + \, \gamma )}{cos \, (\alpha  \, - \, \gamma )}$$ 
on changing to sin and cos 
$$\therefore  \, \sin \, 2\beta  \, = \, \dfrac{2 \, \tan \, \beta }{1 \, + \, \tan^2 \, \beta } \, = \, \dfrac{2 \, \sin \, (\alpha  \, + \, \gamma ) \, \cos \, (\alpha  \, -\, \gamma )}{\cos^2 \, (\alpha  \, - \, \gamma ) \, + \, \sin^2 \, (\alpha  \, + \, \gamma )}$$ 
$$= \, \dfrac{\sin \, 2\alpha  \, + \, \sin \,2\gamma }{\sin^2 \, (\alpha  \, + \, \gamma ) \, + \, 1 \, - \, \sin^2(\alpha  \, - \, \gamma )}$$ 
$$= \, \dfrac{\sin \, 2\alpha  \, + \, \sin \,2\gamma }{1 \, + \, \sin \, 2\alpha  \, \sin \, 2\gamma }$$ by $$\sin^2 \, A \, - \, \sin^2 \, B$$ 
using formula of $$\sin^2 \, A \, - \, \sin^2 \, B$$