1. Let the function $f(x)$ have the domain $\mathbf{R}$, and for any real number $x \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right), f(\tan x)=\sin 2 x$. Then the maximum value of $f(2 \sin x)$ is $(\quad)$.
(A) 0
(B) $\frac{1}{2}$
(C) $\frac{\sqrt{2}}{2}$
(D) 1
-,1.D.
From $f(\tan x)=\frac{2 \tan x}{1+\tan ^{2} x}$, we know $f(x)=\frac{2 x}{1+x^{2}}$. Therefore, $f(2 \sin x)=\frac{4 \sin x}{1+4 \sin ^{2} x} \leq 1$.

When $\sin x=\frac{1}{2}$, the equality holds.
Thus, the maximum value of $f(2 \sin x)$ is 1.