10. Let $\alpha, \beta \in\left(0, \frac{\pi}{2}\right)$, find the maximum value of $A=\frac{\left(1-\sqrt{\tan \frac{\alpha}{2} \tan \frac{\beta}{2}}\right)^{2}}{\cot \alpha+\cot \beta}$.
10. Solution: Let $\tan \frac{\alpha}{2}=p, \tan \frac{\beta}{2}=q$, given $\alpha, \beta \in\left(0, \frac{\pi}{2}\right)$, we know $p, q \in(0,1)$,

Thus, $A=\frac{(1-\sqrt{p q})^{2}}{\frac{1-p^{2}}{2 p}+\frac{1-q^{2}}{2 q}}=\frac{2 p q(1-\sqrt{p q})^{2}}{(p+q)(1-p q)} \leqslant \frac{\sqrt{p q}(1-\sqrt{p q})^{2}}{1-p q}$,
Let $\sqrt{p q}+1=u \in(0,2)$, substituting we get $A \leqslant \frac{(u-1)(2-u)}{u}=3-\left(u+\frac{2}{u}\right) \leqslant 3-2 \sqrt{2}$.
Equality holds if and only if “ $p=q$ and $\sqrt{p q}+1=\sqrt{2}$ ”, i.e., $\tan \frac{\alpha}{2}=\tan \frac{\beta}{2}=\sqrt{2}-1$, hence $\alpha=\beta=\frac{\pi}{4}$, at which $A$ attains its maximum value of $3-2 \sqrt{2}$.