10. (15 points) Through the center $O$ of the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$, draw two perpendicular chords $A C, B D$. Let the eccentric angles of points $A$, $B$ be $\theta_{1}$ and $\theta_{2}$ (here the eccentric angle of $A$ being $\theta_{1}$ means that the coordinates of $A$ are $\left(a \cos \theta_{1}, b \sin \theta_{1}\right)$), find the range of $\left|\cos \left(\theta_{1}-\theta_{2}\right)\right|$.
10. Solution: When $A C, B D$ coincide exactly with the coordinate axes, $\mid \cos \left(\theta_{1} - \theta_{2}\right) \mid=0$; 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 $\varphi_{1}=\varphi_{2} \pm \frac{\pi}{2} + 2 k \pi (k \in \mathbf{Z})$, hence $\tan \varphi_{1} \cdot \tan \varphi_{2}=-1$. From the problem, we know $A \left(b \cos \theta_{1}, a \sin \theta_{1}\right), B\left(b \cos \theta_{2}, a \sin \theta_{2}\right)$, then $\tan \varphi_{1}=\frac{a}{b} \tan \theta_{1}, \tan \varphi_{2}=\frac{a}{b} \tan \theta_{2}$
$$
\begin{array}{l}
\text { Therefore, }\left|\cot \left(\theta_{1}-\theta_{2}\right)\right|=\left|\frac{1+\tan \theta_{1} \cdot \tan \theta_{2}}{\tan \theta_{1}-\tan \theta_{2}}\right| \\
=\left|\frac{a^{2}+b^{2} \tan \varphi_{1} \cdot \tan \varphi_{2}}{a b\left(\tan \varphi_{1}-\tan \varphi_{2}\right)}\right| \\
=\frac{a^{2}-b^{2}}{a b} \cdot \frac{1}{\left|-\cot \varphi_{2}-\tan \varphi_{2}\right|} \leqslant \frac{a^{2}-b^{2}}{2 a b} \\
\left|\cos \left(\theta_{1}-\theta_{2}\right)\right|=\sqrt{1-\frac{1}{1+\cot ^{2}\left(\theta_{1}-\theta_{2}\right)}} \\
\leqslant \sqrt{1-\frac{1}{1+\left(\frac{a^{2}-b^{2}}{2 a b}\right)^{2}}}=\frac{a^{2}-b^{2}}{a^{2}+b^{2}} .
\end{array}
$$

The equality condition is $\left|\tan \varphi_{2}\right|=1$, i.e., the inclination angle of $B D$ is $\frac{\pi}{4}$ or $\frac{3 \pi}{4}$, hence $0 \leqslant\left|\cos \left(\theta_{1}-\theta_{2}\right)\right| \leqslant \frac{a^{2}-b^{2}}{a^{2}+b^{2}}$.