For example, if $0<\alpha<\frac{\pi}{2}, 0<\beta<\frac{\pi}{2}$, find the extremum of $\frac{1}{\cos ^{2} \alpha}+\frac{1}{\sin ^{2} \alpha \sin ^{2} \beta \cos ^{2} \beta}$, and the corresponding values of $\alpha$ and $\beta$.
$$
\begin{array}{l}
\geqslant 2 \sqrt{\sin ^{2} \alpha} \cos ^{2} \alpha \sin ^{2} \beta \cos ^{2} \beta \\
=\frac{8}{\sin 2 \alpha \sin 2 \beta} \geqslant 8 \text {, it is easy to list the above formula to get the minimum value } \\
\end{array}
$$
The condition for 8 is $\sin 2 \alpha=\sin 2 \beta=1$, at this time $\alpha=\beta=\frac{\pi}{4}$. Note: This answer is incorrect.

Because when $\alpha=\beta=\frac{\pi}{4}$, $\frac{1}{\cos ^{2} \alpha}+$
$$
\overline{\sin } \frac{1}{\sin ^{2} \beta \cos ^{2} \beta}=2+8=10 \neq 8 \text {. As for }
$$

where the mistake lies, it is left for the reader to consider.
Since the variables in the problem are "symmetric," but the conclusion does not conform to the above "rule," it indicates that this "rule" is not reliable.
The correct solution is:
$$
\begin{array}{l}
\frac{1}{\cos ^{2} \alpha}+\frac{1}{\sin ^{2} \alpha \sin ^{2} \beta \cos ^{2} \beta}=\frac{1}{\cos ^{2} \alpha} \\
+\frac{4}{\sin ^{2} \alpha \sin ^{2} 2 \beta} \stackrel{1}{\cos ^{2} \alpha}+\frac{4}{\sin ^{2} \alpha} \\
=1+\operatorname{tg}^{2} \alpha+4\left(1+\operatorname{ctg}^{2} \alpha\right)=5+\operatorname{tg}^{2} \alpha \\
+4 \cot \approx^{2} \in \geqslant 5+2 \sqrt{1 g^{2}} \bar{c} \cdot 1 \operatorname{cog}^{2} \alpha=5+2 \times 2 \\
=9 \text {, so } \frac{1}{\cos ^{2} \alpha}+\frac{1}{\sin ^{2} \alpha \sin ^{2} \beta \cos ^{2} \beta} \\
\geqslant 9 \text {. } \\
\end{array}
$$
(1) The equality in (1) holds only when $\sin 2 \beta=1$, i.e., $\beta=\frac{\pi}{4}$; (2) The equality in (2) holds only when $\operatorname{tg}^{2} \alpha=4 \operatorname{ctg}^{2} \alpha$, i.e., $\operatorname{tg}^{2} \alpha=2, \quad \operatorname{tg} \alpha=\sqrt{2}, \quad \alpha=\operatorname{arctg} \sqrt{2}$.
$\therefore$ When $\boldsymbol{\beta}=\frac{\pi}{4}, \alpha=\operatorname{arctg} \sqrt{2}$, the extremum in the original expression is 9, and at this time $\alpha \neq \beta$.