32. Let $\alpha, \beta \in\left(0, \frac{\pi}{2}\right)$, find the maximum value of $A=\frac{\left(1-\sqrt{\tan \frac{\alpha}{2} \tan \frac{\beta}{2}}\right)^{2}}{\cot \alpha+\cot \beta}$.
32. Let $x=\tan \frac{\alpha}{2}, y=\tan \frac{\beta}{2}, x, y \in(0,1)$, then $\cot \alpha+\cot \beta=\frac{1-x^{2}}{2 x}+\frac{1-y^{2}}{2 y} =\frac{(x+y)(1-x y)}{2 x y}$, so, $A=\frac{2 x y(1-\sqrt{x y})^{2}}{(x+y)(1-x y)}=\frac{2 x y(1-\sqrt{x y})}{(x+y)(1+\sqrt{x y})} \leqslant$ $\frac{2 x y(1-\sqrt{x y})}{2 \sqrt{x y}(1+\sqrt{x y})}=\frac{\sqrt{x y}(1-\sqrt{x y})}{1+\sqrt{x y}}$

Let $t=\sqrt{x y}, t \in(0,1)$, then $A=\frac{t(1-t)}{1+t}=\frac{-(1+t)^{2}+3(1+t)-2}{1+t}=$ $3-\left(1+t+\frac{2}{1+t}\right) \leqslant 3-2 \sqrt{2}$

When and only when $t=\sqrt{2}-1$, i.e., $\tan \frac{\alpha}{2}=\tan \frac{\beta}{2}=\sqrt{2}-1$, the equality holds. Hence $A_{\max }=3-2 \sqrt{2}$.