Example 4 Given that $\alpha$ is an acute angle, $\beta$ is an obtuse angle, and $\sec (\alpha-2 \beta) 、 \sec \alpha 、 \sec (\alpha+2 \beta)$ form an arithmetic sequence, find the value of $\frac{\cos \alpha}{\cos \beta}$.
Solving by transforming the given condition into a trigonometric identity, we get: 
\[
\frac{2}{\cos \alpha} = \frac{1}{\cos (\alpha - 2 \beta)} + \frac{1}{\cos (\alpha + 2 \beta)}, \quad \frac{2}{\cos \alpha} = \frac{\cos (\alpha + 2 \beta) + \cos (\alpha - 2 \beta)}{\cos (\alpha + 2 \beta) \cos (\alpha - 2 \beta)}, \quad \frac{2}{\cos \alpha} = \frac{2 \cos \alpha \cos 2 \beta}{\frac{1}{2}(\cos 2 \alpha + \cos 4 \beta)},
\]
\[
\cos 2 \alpha + \cos 4 \beta = 2 \cos^2 \alpha \cos 2 \beta, \quad \cos 2 \alpha + \cos 4 \beta = (1 + \cos 2 \alpha) \cos 2 \beta, \quad \cos 2 \alpha (1 - \cos 2 \beta) = \cos 2 \beta - \cos 4 \beta,
\]
\[
\cos 2 \alpha (1 - \cos 2 \beta) = \cos 2 \beta - 2 \cos^2 2 \beta + 1, \quad \cos 2 \alpha (1 - \cos 2 \beta) = (1 - \cos 2 \beta)(2 \cos 2 \beta + 1),
\]
\[
\cos 2 \alpha = 2 \cos 2 \beta + 1, \quad \cos^2 \alpha = 2 \cos^2 \beta.
\]
Given that \(\alpha\) is an acute angle and \(\beta\) is an obtuse angle, we get:
\[
\frac{\cos \alpha}{\cos \beta} = -\sqrt{2}.
\]