Answer
$$\frac{\sin(x-y)}{\sin(x+y)}=\frac{\tan x-\tan y}{\tan x+\tan y}$$
The equation has been verified to be an identity.
Work Step by Step
$$\frac{\sin(x-y)}{\sin(x+y)}=\frac{\tan x-\tan y}{\tan x+\tan y}$$
About the right side, we can use the identity $\tan x=\frac{\sin x}{\cos x}$ $$\frac{\tan x-\tan y}{\tan x+\tan y}$$ $$=\frac{\frac{\sin x}{\cos x}-\frac{\sin y}{\cos y}}{\frac{\sin x}{\cos x}+\frac{\sin y}{\cos y}}$$ $$=\frac{\frac{\sin x\cos y-\cos x\sin y}{\cos x\cos y}}{\frac{\sin x\cos y+\cos x\sin y}{\cos x\cos y}}$$ $$=\frac{\sin x\cos y-\cos x\sin y}{\sin x\cos y+\cos x\sin y}$$ $$=\frac{\sin (x-y)}{\sin(x+y)}$$
(identities of sine of a sum and a difference)
Therefore, the equation is an identity.
