$9 \cdot 23$ The largest of the following four numbers is
(A) $\operatorname{tg} 48^{\circ}+\operatorname{ctg} 48^{\circ}$.
(B) $\sin 48^{\circ}+\cos 48^{\circ}$.
(C) $\operatorname{tg} 48^{\circ}+\cos 48^{\circ}$.
(D) $\operatorname{ctg} 48^{\circ}+\sin 48^{\circ}$.
(Chinese Junior High School Mathematics League, 1988)
[Solution] Since $0(C)>(D)>(B)$, when $45^{\circ}(C)=(D)>(B)$, when $\alpha=45^{\circ}$;
(A) $>(D)>(C)>(B)$, when $0^{\circ}<\alpha<45^{\circ}$, $D C, B B^{\prime}>O D$,
we have $A A^{\prime}+B B^{\prime}>B B^{\prime}+D C>D C+O D$,

and $A A^{\prime}+B B^{\prime}>A A^{\prime}+O D>D C+O D$.
That is, $\operatorname{tg} \alpha+\operatorname{ctg} \alpha>\operatorname{ctg} \alpha+\sin \alpha>\sin \alpha+\cos \alpha$,

and $\operatorname{tg} \alpha+\operatorname{ctg} \alpha>\operatorname{tg} \alpha+\cos \alpha>\sin \alpha+\cos \alpha$.
Moreover, by $\triangle B^{\prime} E C \backsim \triangle C F A^{\prime}$, and $C B^{\prime}\operatorname{ctg} \alpha+\sin \alpha$. Thus, $(A)>(C)>(D)>(B)$.