For $\theta<\pi$, find the maximum real number $\theta$ such that
$$
\prod_{k=0}^{10} \cos 2^{k} \theta \neq 0, \prod_{k=0}^{10}\left(1+\frac{1}{\cos 2^{k} \theta}\right)=1
$$
To make the equation hold, $\theta$ cannot be an integer multiple of $\pi$. Otherwise, $\sin \theta=0, \cos \theta= \pm 1$.
$$
\begin{array}{l}
=\frac{z+z^{-1}}{z^{21}+z^{-2^{11}}} \cdot \frac{z^{z^{11}}-z^{-2^{11}}}{z-z^{-1}}=\frac{\tan 2^{10} \theta}{\tan \frac{\theta}{2}}=1 \text {. } \\
\end{array}
$$

Since the period of $\tan x$ is $\pi$, and it is injective within any given period, we have:
$$
\frac{\theta}{2}+l \pi=1024 \theta(l \in \mathbf{Z}) \Rightarrow \theta=\frac{2 l \pi}{2047} .
$$

When $l=1023$, $\theta$ reaches its maximum value $\frac{2046 \pi}{2047}$.