Answer
$\dfrac{\sin 3 x+\cos 3x}{\cos x-\sin x}=1+4 \sin x\cos x$
Work Step by Step
Consider LHS: $\dfrac{\sin 3 x+\cos 3x}{\cos x-\sin x}=\dfrac{\sin x(\cos^2 x-\sin^2 x)+2\cos^2 x\sin x+\cos x(\cos^2 x-\sin^2 x)}{\cos x-\sin x}$
or, $\dfrac{\sin x(\cos^2 x-\sin^2 x)+2\cos^2 x\sin x+\cos x(\cos^2 x-\sin^2 x)}{\cos x-\sin x}=(\cos x+\sin x)^2+2\sin x\cos x$
or, $\cos^2 x +\sin^2 x+2 \sin x\cos x+2 \sin x\cos x=1+4 \sin x\cos x$
Thus, $\dfrac{\sin 3 x+\cos 3x}{\cos x-\sin x}=1+4 \sin x\cos x$
Hence, the result has been proved.