1. The range of the function $f(x)=\left(|\cot x|+\frac{1}{\sin x}\right)^{2}$ $\left(0<|x|<\frac{\pi}{2}\right)$ is ( ).
(A) $\mathbf{R}_{+}$
(B) $\mathbf{R}_{+} \backslash\{1\}$
(C) $\mathbf{R}_{+} \backslash\{\sqrt{2}\}$
(D) $\mathbf{R}_{+} \backslash\{1, \sqrt{2}\}$
- 1.B.

When $0<x<\frac{\pi}{2}$,
$$
\begin{array}{l}
f(x)=\left(\cot x+\frac{1}{\sin x}\right)^{2}=\left(\frac{1+\cos x}{\sin x}\right)^{2} \\
=\left(\frac{2 \cos ^{2} \frac{x}{2}}{2 \sin \frac{x}{2} \cdot \cos \frac{x}{2}}\right)^{2}=\cot ^{2} \frac{x}{2} \in(1,+\infty) ;
\end{array}
$$

When $-\frac{\pi}{2}<x<0$,
$$
\begin{array}{l}
f(x)=\left(-\cot x+\frac{1}{\sin x}\right)^{2}=\left(\frac{-\cos x+1}{\sin x}\right)^{2} \\
=\tan ^{2} \frac{x}{2} \in(0,1) .
\end{array}
$$

Therefore, the range of $f(x)$ is $\mathbf{R}_{+} \mid\{1\}$.