11. 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
II, 11.0.
From the given, we have
$$
\begin{array}{l}
\frac{\sin (x+y)}{x+y}=\frac{\sin (x+y)}{\tan x+\tan y} \\
=\frac{\sin (x+y)}{\frac{\sin x}{\cos x}+\frac{\sin y}{\cos y}}=\cos x \cdot \cos y .
\end{array}
$$

Similarly, $\frac{\sin (x-y)}{x-y}=\cos x \cdot \cos y$. Therefore, $\frac{\sin (x+y)}{x+y}-\frac{\sin (x-y)}{x-y}=0$.