Answer
$$\frac{\sin(x+y)}{\cos(x-y)}=\frac{\cot x+\cot y}{1+\cot x\cot y}$$
The equation is an identity, as proved below.
Work Step by Step
$$\frac{\sin(x+y)}{\cos(x-y)}=\frac{\cot x+\cot y}{1+\cot x\cot y}$$
We start from the right side in this exercise.
$$X=\frac{\cot x+\cot y}{1+\cot x\cot y}$$
- For $\cot x$ and $\cot y$, we can apply the following identity:
$$\cot\theta=\frac{\cos\theta}{\sin\theta}$$
$$X=\frac{\frac{\cos x}{\sin x}+\frac{\cos y}{\sin y}}{1+\frac{\cos x}{\sin x}\frac{\cos y}{\sin y}}$$
$$X=\frac{\frac{\cos x\sin y+\sin x\cos y}{\sin x\sin y}}{1+\frac{\cos x\cos y}{\sin x\sin y}}$$
$$X=\frac{\frac{\cos x\sin y+\sin x\cos y}{\sin x\sin y}}{\frac{\sin x\sin y+\cos x\cos y}{\sin x\sin y}}$$
$$X=\frac{\sin x\cos y+\cos x\sin y}{\cos x\cos y+\sin x\sin y}$$
Now recall that
$$\sin(x+y)=\sin x\cos y+\cos x\sin y$$
$$\cos(x-y)=\cos x\cos y+\sin x\sin y$$
Therefore, $$X=\frac{\sin(x+y)}{\cos(x-y)}$$
Thus, $$\frac{\sin(x+y)}{\cos(x-y)}=\frac{\cot x+\cot y}{1+\cot x\cot y}$$
As 2 sides are equal, the equation is an identity.
