Answer
$$\frac{\cos(x+y)+\cos(x-y)}{\sin(x-y)+\sin(x+y)}=\cot x$$
The equation is an identity.
Work Step by Step
$$\frac{\cos(x+y)+\cos(x-y)}{\sin(x-y)+\sin(x+y)}=\cot x$$
We should examine from the left side.
$$A=\frac{\cos(x+y)+\cos(x-y)}{\sin(x-y)+\sin(x+y)}$$
As in here, the application of 4 following identities is essential:
$$\cos(A+B)=\cos A\cos B-\sin A\sin B$$
$$\cos(A-B)=\cos A\cos B+\sin A\sin B$$
$$\sin(A+B)=\sin A\cos B+\cos A\sin B$$
$$\sin(A-B)=\sin A\cos B-\cos A\sin B$$
That means $$A=\frac{\cos x\cos y-\sin x\sin y+\cos x\cos y+\sin x\sin y}{\sin x\cos y-\cos x\sin y+\sin x\cos y+\cos x\sin y}$$
$$A=\frac{(\cos x\cos y+\cos x\cos y)+(-\sin x\sin y+\sin x\sin y)}{(\sin x\cos y+\sin x\cos y)+(-\cos x\sin y+\cos x\sin y)}$$
$$A=\frac{2\cos x\cos y}{2\sin x\cos y}$$
$$A=\frac{\cos x}{\sin x}$$
$$A=\cot x\hspace{1cm}\text{(Quotient Identity)}$$
The equation has thus been verified to be an identity.
