$13 \cdot 52$ function $f(x)=(1+\sec x) \operatorname{ctg}^{2} \frac{x}{2}\left(0 \leqslant x \leqslant \frac{\pi}{2}\right)$
(A) has a minimum value, but no maximum value.
(B) has a maximum value, but no minimum value.
(C) has both a minimum value and a maximum value.
(D) has neither a minimum value nor a maximum value.
(2nd "Hope Cup" National Mathematics Invitational Competition, 1991)
［Solution］Notice
$$
\begin{array}{l} 
f(x)=\frac{1+\cos x}{\cos x} \cdot \cot^{2} \frac{x}{2}=\frac{\frac{1-\tan^{2} \frac{x}{2}}{1+\tan^{2} \frac{x}{2}}+1}{\frac{1-\tan^{2} \frac{x}{2}}{1+\tan^{2} \frac{x}{2}}} \cdot \cot^{2} \frac{x}{2} \\
=\frac{2}{\left(1-\tan^{2} \frac{x}{2}\right) \tan^{2} \frac{x}{2}}, \\
\text { and } 0 \leqslant\left(1-\tan^{2} \frac{x}{2}\right) \tan^{2} \frac{x}{2} \leqslant\left(\frac{1-\tan^{2} \frac{x}{2}+\tan^{2} \frac{x}{2}}{2}\right)^{2}=\frac{1}{4},
\end{array}
$$

Therefore, $f(x) \geqslant 8$. Thus, $f_{\min }(x)=8$, but $f(x)$ has no maximum value.
Hence, the correct choice is $(A)$.
Note The original proposition has no interval restriction. In this case,
by $f(x)=\frac{1+\cos x}{\cos x} \cdot \frac{1+\cos x}{1-\cos x}=\frac{(1+\cos x)^{2}}{\cos x(1-\cos x)}$,
when $x>\frac{\pi}{2}, x \rightarrow \frac{\pi}{2}$, $f(x) \rightarrow-\infty$,
when $x<\frac{\pi}{2}, x \rightarrow \frac{\pi}{2}$, $f(x) \rightarrow+\infty$,
thus, $f(x)$ has neither a maximum value nor a minimum value.