If frac{tan (alpha+beta-gamma)}{tan (alpha-be | vedclass

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(A) Given $\frac{	an (\alpha+\beta-\gamma)}{	an (\alpha-\beta+\gamma)}=\frac{	an \gamma}{	an \beta}$.Applying componendo and dividendo:$\frac{	an

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(A) Given $\frac{	an (\alpha+\beta-\gamma)}{	an (\alpha-\beta+\gamma)}=\frac{	an \gamma}{	an \beta}$.Applying componendo and dividendo:$\frac{	an (\alpha+\beta-\gamma)+	an (\alpha-\beta+\gamma)}{	an (\alpha+\beta-\gamma)-	an (\alpha-\beta+\gamma)}=\frac{	an \gamma+	an \beta}{	an \gamma-	an \beta}$.Using $	an A + 	an B = \frac{\sin(A+B)}{\cos A \cos B}$ and $	an A - 	an B = \frac{\sin(A-B)}{\cos A \cos B}$,we get:$\frac{\sin(2\alpha)}{\sin(2\beta-2\gamma)} = \frac{\sin(\gamma+\beta)}{\sin(\gamma-\beta)}$.$\sin(2\alpha) = \frac{\sin(\gamma+\beta) \sin(2\beta-2\gamma)}{\sin(\gamma-\beta)} = \frac{\sin(\gamma+\beta) \cdot 2 \sin(\beta-\gamma) \cos(\beta-\gamma)}{-(\sin(\beta-\gamma))} = -2 \sin(\beta+\gamma) \cos(\beta-\gamma)$.Using $2 \sin A \cos B = \sin(A+B) + \sin(A-B)$:$\sin(2\alpha) = -(\sin(2\beta) + \sin(2\gamma))$.Therefore,$\sin 2\alpha + \sin 2\beta + \sin 2\gamma = 0$.

If $\frac{\tan (\alpha+\beta-\gamma)}{\tan (\alpha-\beta+\gamma)}=\frac{\tan \gamma}{\tan \beta}$ and $\beta \neq \gamma$,then the value of $\sin 2 \alpha+\sin 2 \beta+\sin 2 \gamma$ is

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