A) 1
B) - 1
C) 0
D) None of these
Correct Answer: B
Solution :
We have \[\sin x+\sin y=3\,(\cos y-\cos x)\] \[\Rightarrow \,\sin x+3\cos x=3\cos y-\sin y\] ?..(i) \[\Rightarrow \,r\cos \,(x-\alpha )=r\cos \,(y+\alpha ),\] where \[r=\sqrt{10},\,\tan \alpha =\frac{1}{3}\] \[\Rightarrow \,x-\alpha =\pm (y+\alpha )\,\Rightarrow \,x=-y\] or \[x+y=2\alpha \] Clearly, \[x=-y\]satisfies (i); \[\therefore \ \frac{\sin \,3x}{\sin \,3y}=\frac{-\sin \,3y}{\sin \,3y}=-1\].
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