## Trigonometry (11th Edition) Clone

$$\frac{\cos(x+y)+\cos(x-y)}{\sin(x-y)+\sin(x+y)}=\cot x$$ The equation is an identity.
$$\frac{\cos(x+y)+\cos(x-y)}{\sin(x-y)+\sin(x+y)}=\cot x$$ 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.