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,we get: $\sin x + 3\cos x = 3\cos y - \sin y$.....$(i)$Divide both sides by $\s

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$-1$

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(B) Given: $\sin x + \sin y = 3(\cos y - \cos x)$Rearranging the terms,we get: $\sin x + 3\cos x = 3\cos y - \sin y$.....$(i)$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 $\sin \alpha = \frac{3}{\sqrt{10}}$,so $	an \alpha = 3$. Then the equation becomes:$\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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