1. Given $\theta \in\left[\frac{5 \pi}{4}, \frac{3 \pi}{2}\right]$, then $\sqrt{1-\sin 2 \theta}-\sqrt{1+\sin 2 \theta}$ can be simplified to
A. $2 \sin \theta$
B. $-2 \sin \theta$
C. $-2 \cos \theta$
D. $2 \cos \theta$
Solution: Since $\theta \in\left[\frac{5 \pi}{4}, \frac{3 \pi}{2}\right]$, we have $\sqrt{1-\sin 2 \theta}-\sqrt{1+\sin 2 \theta}=|\cos \theta-\sin \theta|-|\cos \theta+\sin \theta|$ $=2 \cos \theta$. The correct answer is $\mathrm{D}$.