If tan beta = frac{tan alpha + tan gamma}{1 + | vedclass

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(A) Given,$	an \beta = \frac{	an \alpha + 	an \gamma}{1 + 	an \alpha 	an \gamma} = \frac{\frac{\sin \alpha}{\cos \alpha} + \frac{\sin \gamma}{\co

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$\sin 2 \beta$

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(A) Given,$	an \beta = \frac{	an \alpha + 	an \gamma}{1 + 	an \alpha 	an \gamma} = \frac{\frac{\sin \alpha}{\cos \alpha} + \frac{\sin \gamma}{\cos \gamma}}{1 + \frac{\sin \alpha}{\cos \alpha} \cdot \frac{\sin \gamma}{\cos \gamma}}$$= \frac{\sin \alpha \cos \gamma + \sin \gamma \cos \alpha}{\cos \alpha \cos \gamma + \sin \alpha \sin \gamma} = \frac{\sin(\alpha + \gamma)}{\cos(\alpha - \gamma)} \dots (1)$Now,consider the expression $E = \frac{\sin 2 \alpha + \sin 2 \gamma}{1 + \sin 2 \alpha \sin 2 \gamma} = \frac{2 \sin(\alpha + \gamma) \cos(\alpha - \gamma)}{1 + (2 \sin \alpha \cos \alpha)(2 \sin \gamma \cos \gamma)}$$= \frac{2 \sin(\alpha + \gamma) \cos(\alpha - \gamma)}{\cos^2(\alpha - \gamma) + \sin^2(\alpha + \gamma)} = \frac{2 \frac{\sin(\alpha + \gamma)}{\cos(\alpha - \gamma)}}{1 + \left(\frac{\sin(\alpha + \gamma)}{\cos(\alpha - \gamma)}ight)^2}$Using equation $(1)$,this is $\frac{2 	an \beta}{1 + 	an^2 \beta} = \sin 2 \beta$.

If $\tan \beta = \frac{\tan \alpha + \tan \gamma}{1 + \tan \alpha \tan \gamma}$,then $\frac{\sin 2 \alpha + \sin 2 \gamma}{1 + \sin 2 \alpha \sin 2 \gamma} = $

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