9.30 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) $(\cos \alpha)^{\sin \alpha}<(\cos \alpha)^{\cos \alpha}<(\sin \alpha)^{\cos \alpha}$.
(China High School Mathematics Competition, 1990)
[Solution] $\because \quad \alpha \in\left(\frac{\pi}{4}, \frac{\pi}{2}\right), \quad \therefore \quad 0<\cos \alpha<\frac{\sqrt{2}}{2}<\sin \alpha<1$. Therefore, $(\cos \alpha)^{\sin \alpha}<(\cos \alpha)^{\cos \alpha},(\cos \alpha)^{\cos \alpha}<(\sin \alpha)^{\cos \alpha}$. Hence, the correct choice is $(D)$.