## Trigonometry (11th Edition) Clone

$$\frac{\sin(x+y)}{\cos(x-y)}=\frac{\cot x+\cot y}{1+\cot x\cot y}$$ After showing that the left side is equal to the right side, the fact that the equation is an identity is verified.
$$\frac{\sin(x+y)}{\cos(x-y)}=\frac{\cot x+\cot y}{1+\cot x\cot y}$$ The left side would be examined first. $$X=\frac{\sin(x+y)}{\cos(x-y)}$$ We apply here the sum identity for sines and difference identity for cosines: $$\sin(A+B)=\sin A\cos B+\cos A\sin B$$ $$\cos(A-B)=\cos A\cos B+\sin A\sin B$$ Therefore, $$X=\frac{\sin x\cos y+\cos x\sin y}{\cos x\cos y+\sin x\sin y}$$ Now we examine the right side. $$Y=\frac{\cot x+\cot y}{1+\cot x\cot y}$$ $$Y=\frac{\frac{\cos x}{\sin x}+\frac{\cos y}{\sin y}}{1+\frac{\cos x}{\sin x}\frac{\cos y}{\sin y}}$$ $$Y=\frac{\frac{\cos x\sin y+\cos y\sin x}{\sin x\sin y}}{1+\frac{\cos x\cos y}{\sin x\sin y}}$$ $$Y=\frac{\frac{\cos x\sin y+\cos y\sin x}{\sin x\sin y}}{\frac{\sin x\sin y+\cos x\cos y}{\sin x\sin y}}$$ $$Y=\frac{\cos x\sin y+\cos y\sin x}{\sin x\sin y+\cos x\cos y}$$ $$Y=\frac{\sin x\cos y+\cos x\sin y}{\cos x\cos y+\sin x\sin y}$$ Thus, $X=Y$ The equation is verified to be an identity.