1. Real numbers $x, y$ satisfy $\tan x=x, \tan y=y$, and $|x| \neq|y|$, then the value of $\frac{\sin (x+y)}{x+y}-\frac{\sin (x-y)}{x-y}$ is . $\qquad$
1. 0 Detailed Explanation: From the problem, we get $\frac{\sin (x+y)}{x+y}=\frac{\sin (x+y)}{\tan x+\tan y}=\frac{\sin (x+y)}{\frac{\sin x}{\cos x}+\frac{\sin y}{\cos y}}=\cos x \cdot \cos y$, similarly $\frac{\sin (x-y)}{x-y}$ $=\cos x \cdot \cos y$, therefore $\frac{\sin (x+y)}{x+y}-\frac{\sin (x-y)}{x-y}=0$.