Example 9 (64th Putnam Mathematical Competition Problem) Find the minimum value of the following algebraic expression: $|\sin x+\cos x+\tan x+\cot x+\sec x+\csc x|$, where $x$ is a real number.
Solve $f(x)=\sin x+\cos x+\tan x+\cot x+\sec x+\csc x$
$$
=\sin x+\cos x+\frac{1}{\sin x \cos x}+\frac{\sin x+\cos x}{\sin x \cos x} \text {. }
$$

Let $t=\sin x+\cos x$, then $\sin x \cos x=\frac{t^{2}-1}{2}$, here $|t| \leqslant \sqrt{2}$, and $|t| \neq 1$.
$$
f(x)=t+\frac{2}{t^{2}-1}+\frac{2 t}{t^{2}-1}=t+\frac{2}{t-1}=t-1+\frac{2}{t-1}+1 .
$$

When $t>1$, $t-1+\frac{2}{t-1} \geqslant 2 \sqrt{2}$, at this time $f(x) \geqslant 1+2 \sqrt{2}$ (in fact, the equality cannot be achieved, because the equality holds when $t=1+\sqrt{2}$, which exceeds the range of $t$).

When $t<1$, $-f(x)=-1+(1-t)+\frac{2}{1-t} \geqslant-1+2 \sqrt{2}$ (the equality is achieved when $t=1-\sqrt{2}$). These discussions show that the minimum value of $|f(x)|$ is $2 \sqrt{2}-1$.