Example 8 Given that the three interior angles $A, B, C$ of $\triangle ABC$ satisfy $A+C=2B$, find the range of $\sec A+\sec C$.

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From $A+C=2B$ we know $3B=2B+B=A+C+B=\pi$, hence $B=\frac{\pi}{3}$. We can set $A=\frac{\pi}{3}+x, C=\frac{\pi}{3}-x, x \in\left(-\frac{\pi}{3}, \frac{\pi}{3}\right)$ and $x \neq \pm \frac{\pi}{6}$.
$$
\begin{aligned}
\sec A+\sec C & =\frac{1}{\cos A}+\frac{1}{\cos C}=\frac{1}{\cos \left(\frac{\pi}{3}+x\right)}+\frac{1}{\cos \left(\frac{\pi}{3}-x\right)}=\frac{\cos \left(\frac{\pi}{3}-x\right)+\cos \left(\frac{\pi}{3}+x\right)}{\cos \left(\frac{\pi}{3}+x\right) \cos \left(\frac{\pi}{3}-x\right)} \\
& =\frac{2 \cos \frac{\pi}{3} \cos x}{\cos ^{2} \frac{\pi}{3} \cos ^{2} x-\sin ^{2} \frac{\pi}{3} \sin ^{2} x}=\frac{4 \cos x}{\cos ^{2} x-3 \sin ^{2} x}=\frac{4 \cos x}{4 \cos ^{2} x-3} .
\end{aligned}
$$

Let $t=\cos x$, then $x \in\left(\frac{1}{2}, 1\right]$, and $t \neq \frac{\sqrt{3}}{2}$.
$$
\sec A+\sec C=\frac{4 t}{4 t^{2}-3}=\frac{4}{4 t-\frac{3}{t}} \text {. }
$$

Since the function $f(x)=4 x-\frac{3}{x}$ is monotonically increasing in $(-\infty, 0)$ and $(0,+\infty)$.
Therefore, the range of $\sec A+\sec C$ is $(-\infty,-1) \cup[4,+\infty)$.