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); \[\therefore \ \frac{\sin \,3X}{\sin \,3Y}=\frac{-\sin \,3y}{\sin \,3y}=-1\].