Answer
$$\frac{\cos(x+y)+\cos(y-x)}{\sin(x+y)-\sin(y-x)}=\cot x$$
After proving 2 sides are equal to each other, we can state that the equation is an identity.
Work Step by Step
$$\frac{\cos(x+y)+\cos(y-x)}{\sin(x+y)-\sin(y-x)}=\cot x$$
From the left first:
$$X=\frac{\cos(x+y)+\cos(y-x)}{\sin(x+y)-\sin(y-x)}$$
- A recount of all the identities here:
$$\cos(x+y)=\cos x\cos y-\sin x\sin y$$
$$\cos(y-x)=\cos y\cos x+\sin y\sin x=\cos x\cos y+\sin x\sin y$$
$$\sin(x+y)=\sin x\cos y+\cos x\sin y$$
$$\sin(y-x)=\sin y\cos x-\cos y\sin x=\cos x\sin y-\sin x\cos y$$
Therefore:
- Numerator: $$\cos(x+y)+\cos(y-x)=\cos x\cos y-\sin x\sin y+\cos x\cos y+\sin x\sin y$$
$$\cos(x+y)+\cos(y-x)=2\cos x\cos y$$
- Denominator: $$\sin(x+y)-\sin(y-x)=\sin x\cos y+\cos x\sin y-(\cos x\sin y-\sin x\cos y)$$
$$\sin(x+y)-\sin(y-x)=\sin x\cos y+\cos x\sin y-\cos x\sin y+\sin x\cos y$$
$$\sin(x+y)-\sin(y-x)=2\sin x\cos y$$
Apply back to $X$:
$$X=\frac{2\cos x\cos y}{2\sin x\cos y}$$
$$X=\frac{\cos x}{\sin x}$$
$$X=\cot x\hspace{1cm}\text{Quotient Identities}$$
Therefore, $$\frac{\cos(x+y)+\cos(y-x)}{\sin(x+y)-\sin(y-x)}=\cot x$$
That makes the equation eligible to be an identity.
