Example 15 Given $\cos \alpha+\cos \beta=2 m, \sin \alpha+\sin \beta=2 n$. Find the value of $\cot \alpha \cot \beta$.
Let $A(\cos \alpha, \sin \alpha), B(\cos \beta, \sin \beta)$, then we know that $A, B$ are on the circle $x^{2}+y^{2}=1$. Let the midpoint of line segment $AB$ be $C(m, n)$, then $m=\frac{1}{2}(\cos \alpha+\cos \beta), n=\frac{1}{2}(\sin \alpha+\sin \beta)$, and $k_{O C}=\frac{n}{m}, k_{A B}=-\frac{m}{n}$. The equation of $AB$ is $y=-\frac{m}{n} x+\frac{m^{2}+n^{2}}{n}$, and substituting into the equation of the circle, we get $\left(1+\frac{m^{2}}{n^{2}}\right) x^{2}-\frac{2 m\left(m^{2}+n^{2}\right)}{n^{2}} x+$ $\frac{\left(n^{2}+m^{2}\right)^{2}-n^{2}}{n^{2}}=0$. Since $\cos \alpha, \cos \beta$ are the two roots of this equation, we have $\cos \alpha \cdot \cos \beta=\frac{\left(n^{2}+m^{2}\right)^{2}-n^{2}}{m^{2}+n^{2}}$. Similarly, $\sin \alpha \cdot \sin \beta=\frac{\left(m^{2}+n^{2}\right)-m^{2}}{m^{2}+n^{2}}$. Therefore, $\cot \alpha \cdot \cot \beta=\frac{\left(m^{2}+n^{2}\right)^{2}-n^{2}}{\left(m^{2}+n^{2}\right)-m^{2}}$ is the desired result.