1. Let real numbers $x, y$ satisfy $y>x>0$, and
$$
\begin{array}{l}
\tan x=-x, \tan y=-y . \\
\text { Then } \frac{\sin (x-y)}{x-y}-\frac{\sin (x+y)}{x+y}+\frac{2 x^{2} \cos x(x \cos x-\sin x)}{2 x+\sin 2 x}
\end{array}
$$

is ( ).
(A) 0
(B) 2
(C) -1
(D) $-\frac{1}{2}$
$-1 . B$.
From the given, we have
$$
\begin{array}{l}
\frac{\sin (x-y)}{x-y}=\frac{\sin (x-y)}{\tan y-\tan x}=-\cos x \cdot \cos y, \\
\frac{\sin (x+y)}{x+y}=-\cos x \cdot \cos y, \\
\frac{2 x^{2} \cos x(x \cos x-\sin x)}{2 x+\sin 2 x} \\
=\frac{-2 x \sin x(-\sin x-\sin x)}{-2 \tan x+2 \sin x \cdot \cos x} \\
=\frac{2 x \sin x}{-\sec x+\cos x}=\frac{-2 x \cos x \cdot \sin x}{1-\cos ^{2} x}=\frac{2 \sin ^{2} x}{\sin ^{2} x}=2 .
\end{array}
$$

Therefore, the answer is (B).