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If $\displaystyle \frac{\tan(\alpha+\beta-\gamma)}{\tan(\alpha-\beta+\gamma)} = \frac{\tan \gamma}{\tan \beta}.$ Then prove that $\sin (\beta-\gamma) = 0$ or $\sin 2 \alpha+\sin 2\beta+\sin 2 \gamma = 0$

$\bf{My\; Try::}$ Let $\beta-\gamma = \delta\;,$ Then $\displaystyle \frac{\tan (\alpha+\delta)}{\tan(\alpha-\delta)} = \frac{\tan \gamma}{\tan \beta}.$

Using Componendo and Dividendo, We get

$$\frac{\tan(\alpha+\delta)+\tan (\alpha-\delta)}{\tan(\alpha+\delta)-\tan (\alpha-\delta)} = \frac{\tan \gamma+\tan \beta}{\tan \gamma-\tan \beta}$$

So $$\frac{\sin (2\alpha)}{\sin (2\delta)} = -\frac{\sin(\beta+\gamma)}{\sin(\beta-\gamma)}$$

Now how can i solve it after that, Help required, Thanks

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$$\frac{\sin (2\alpha)}{\sin 2(\beta-\gamma )} = -\frac{\sin(\beta+\gamma)}{\sin(\beta-\gamma)}$$ $$\sin(\beta-\gamma )\left[\sin2\alpha +2\sin(\beta+\gamma)\cos(\beta-\gamma)\right]=0$$ $$\sin(\beta-\gamma )\left[\sin2\alpha +\sin2\beta +\sin 2\gamma\right]=0$$

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HINT:

The result does not involve $\tan,$ so it deserves elimination.

$$\dfrac{\sin(\alpha+\overline{\beta-\gamma})}{\cos(\alpha+\overline{\beta-\gamma})}\cdot\dfrac{\cos\{\alpha-(\beta-\gamma)\}}{\sin\{\alpha-(\beta-\gamma)\}}=\dfrac{\sin\gamma\cos\beta}{\cos\gamma\sin\beta}$$

Now apply Componendo and Dividendo and then $\sin(A\pm B)$ formula

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