1. Let $\alpha \in\left(\frac{\pi}{4}, \frac{\pi}{2}\right)$, then the order of $(\cos \alpha)^{\cos \alpha},(\sin \alpha)^{\cos \alpha},(\cos \alpha)^{\sin \alpha}$ is
A. $(\cos \alpha)^{\cos \alpha}<(\sin \alpha)^{\cos \alpha}<(\cos \alpha)^{\sin \alpha}$
B. $(\cos \alpha)^{\cos \alpha}<(\cos \alpha)^{\sin \alpha}<(\sin \alpha)^{\cos \alpha}$
C. $(\sin \alpha)^{\cos \alpha}<(\cos \alpha)^{\cos \alpha}<(\cos \alpha)^{\sin \alpha}$
D. $(\cos \alpha)^{\sin \alpha}<(\cos \alpha)^{\cos \alpha}<(\sin \alpha)^{\cos \alpha}$
Notice that when $a \in(0,1)$, the function $y=x^{a}$ is increasing on $(0,+\infty)$, and $y=a^{x}$ is decreasing on $\mathbf{R}$, and $\alpha \in\left(\frac{\pi}{4}, \frac{\pi}{2}\right) \Rightarrow \sin \alpha>\cos \alpha \in(0,1)$, thus $(\sin \alpha)^{\cos \alpha}>(\cos \alpha)^{\cos \alpha}>(\cos \alpha)^{\sin \alpha}$, so the answer is $D$.