1. $\sqrt{1-\sin ^{2} 1997}-\sqrt{1-\cos ^{2} 1997}$ equals ).
(A) $\cos 1997-\sin 1997$
(B) $-\cos 1997-\sin 1997$
(C) $-\cos 1997+\sin 1997$
(D) $\cos 1997+\sin 1997$
- 1. (D).

It is easy to calculate that the terminal side of a 1997 radian angle is in the $\mathrm{N}$-th quadrant. Therefore,
$$
\begin{aligned}
\text { original expression } & =|\cos 1997|-|\sin 1997| \\
& =\cos 1997+\sin 1997 .
\end{aligned}
$$