Answer
$\dfrac{\sin (x+y)-\sin (x-y)}{\cos (x+y)+\cos (x-y)}=\tan y$
Work Step by Step
Consider LHS: $\dfrac{\sin (x+y)-\sin (x-y)}{\cos (x+y)+\cos (x-y)}=\dfrac{2\sin y \cos x}{2\cos x \cos y}$
or, $\dfrac{2\sin y \cos x}{2\cos x \cos y}=\dfrac{\sin y}{\cos x}$
or, $\dfrac{\sin y}{\cos x}=\tan y$
Thus, $\dfrac{\sin (x+y)-\sin (x-y)}{\cos (x+y)+\cos (x-y)}=\tan y$
Hence, the result has been proved.