If sin x + sin y = 3(cos y - cos x), then the | vedclass

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(B) Given: $\sin x + \sin y = 3(\cos y - \cos x)$Rearranging the terms: $\sin x + 3\cos x = 3\cos y - \sin y$ ... $(i)$Divide both sides by $\sqrt{1^2

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$-1$

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(B) Given: $\sin x + \sin y = 3(\cos y - \cos x)$Rearranging the terms: $\sin x + 3\cos x = 3\cos y - \sin y$ ... $(i)$Divide both sides by $\sqrt{1^2 + 3^2} = \sqrt{10}$:$\frac{1}{\sqrt{10}} \sin x + \frac{3}{\sqrt{10}} \cos x = \frac{3}{\sqrt{10}} \cos y - \frac{1}{\sqrt{10}} \sin y$Let $\cos \alpha = \frac{1}{\sqrt{10}}$ and $\sin \alpha = \frac{3}{\sqrt{10}}$,then $	an \alpha = 3$.The equation becomes: $\sin x \cos \alpha + \cos x \sin \alpha = \cos y \sin \alpha - \sin y \cos \alpha$$\sin(x + \alpha) = \sin(\alpha - y)$This implies $x + \alpha = n\pi + (-1)^n(\alpha - y)$.For $n=0$,$x + \alpha = \alpha - y \Rightarrow x = -y$.Substituting $x = -y$ into the expression $\frac{\sin 3x}{\sin 3y}$:$\frac{\sin 3(-y)}{\sin 3y} = \frac{-\sin 3y}{\sin 3y} = -1$.

If $\sin x + \sin y = 3(\cos y - \cos x),$ then the value of $\frac{\sin 3x}{\sin 3y}$ is

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