Three. (20 points) Given the ellipse $\frac{y^{2}}{a^{2}}+\frac{x^{2}}{b^{2}}=1(a>b>0)$, draw two perpendicular chords $A C$ and $B D$ through the center $O$. Let the eccentric angles of points $A$ and $B$ be $\theta_{1}$ and $\theta_{2}$, respectively. Find the range of $\left|\cos \left(\theta_{1}-\theta_{2}\right)\right|$.
When $A C, B D$ coincide with the coordinate axes,
$$
\left|\cos \left(\theta_{1}-\theta_{2}\right)\right|=0 \text {; }
$$

When $A C, B D$ do not coincide with the coordinate axes, let $\angle x O A=\varphi_{1}$, $\angle x O B=\varphi_{2}$, then $\varphi_{1}=\varphi_{2} \pm \frac{\pi}{2}+2 k \pi(k \in \mathbf{Z})$.
Therefore, $\tan \varphi_{1} \cdot \tan \varphi_{2}=-1$.
From the problem, we know that $A\left(b \cos \theta_{1}, a \sin \theta_{1}\right), B\left(b \cos \theta_{2}\right.$, $a \sin \dot{\theta}_{2}$ ), then $\tan \varphi_{1}=\frac{a}{b} \tan \theta_{1}, \tan \varphi_{2}=\frac{a}{b} \tan \theta_{2}$.
Thus, $\left|\cot \left(\theta_{1}-\theta_{2}\right)\right|=\left|\frac{1+\tan \theta_{1} \cdot \tan \theta_{2}}{\tan \theta_{1}-\tan \theta_{2}}\right|$
$$
\begin{array}{l}
=\left|\frac{1+\frac{b^{2}}{a^{2}} \tan \varphi_{1} \cdot \tan \varphi_{2}}{\frac{b}{a}\left(\tan \varphi_{1}-\tan \varphi_{2}\right)}\right| \\
=\frac{a^{2}-b^{2}}{a b} \cdot \frac{1}{1-\cot \varphi_{2}-\tan \varphi_{2} \mid} \\
=\frac{a^{2}-b^{2}}{a b} \cdot \frac{1}{\left|\cot \varphi_{2}\right|+\left|\tan \varphi_{2}\right|} \leqslant \frac{a^{2}-b^{2}}{2 a b} .
\end{array}
$$

Therefore, $\left|\cos \left(\theta_{1}-\theta_{2}\right)\right|=\sqrt{1-\frac{1}{1+\cot ^{2}\left(\theta_{1}-\theta_{2}\right)}}$
$$
\leqslant \sqrt{1-\frac{1}{1+\left(\frac{a^{2}-b^{2}}{2 a b}\right)^{2}}}=\frac{a^{2}-b^{2}}{a^{2}+b^{2}} \text {. }
$$

When and only when $\left|\cot \varphi_{2}\right|=I \tan \varphi_{2} \mid$, i.e., the inclination angle of $B D$ is $\frac{\pi}{4}$ or $\frac{3 \pi}{4}$, the above inequality holds with equality.
Thus, $0 \leqslant\left|\cos \left(\theta_{1}-\theta_{2}\right)\right| \leqslant \frac{a^{2}-b^{2}}{a^{2}+b^{2}}$.