Five. (25 points) Let $\alpha, \beta \in \left(0, \frac{\pi}{2}\right)$. Find the maximum value of
$$
A=\frac{\left(1-\sqrt{\tan \frac{\alpha}{2} \cdot \tan \frac{\beta}{2}}\right)^{2}}{\cot \alpha+\cot \beta}
$$

(Supplied by Northwest Polytechnical University High School)
Five, from $\cot \alpha+\cot \beta$
$$
\begin{aligned}
&=\frac{1-\tan ^{2} \frac{\alpha}{2}}{2 \tan \frac{\alpha}{2}}+\frac{1-\tan ^{2} \frac{\beta}{2}}{2 \tan \frac{\beta}{2}} \\
&=\frac{\left(\tan \frac{\alpha}{2}+\tan \frac{\beta}{2}\right)\left(1-\tan \frac{\alpha}{2} \cdot \tan \frac{\beta}{2}\right)}{2 \tan \frac{\alpha}{2} \cdot \tan \frac{\beta}{2}},
\end{aligned}
$$

we know
$$
\begin{aligned}
A &=\frac{\left(1-\sqrt{\tan \frac{\alpha}{2} \cdot \tan \frac{\beta}{2}}\right)^{2}}{\cot \alpha+\cot \beta} \\
&=\frac{2 \tan \frac{\alpha}{2} \cdot \tan \frac{\beta}{2}\left(1-\sqrt{\tan \frac{\alpha}{2} \cdot \tan \frac{\beta}{2}}\right)^{2}}{\left(\tan \frac{\alpha}{2}+\tan \frac{\beta}{2}\right)\left(1-\tan \frac{\alpha}{2} \cdot \tan \frac{\beta}{2}\right)} .
\end{aligned}
$$

Let $\tan \frac{\alpha}{2}=x, \tan \frac{\beta}{2}=y$. Then
$$
\begin{aligned}
A &=\frac{2 x y(1-\sqrt{x y})}{(x+y)(1+\sqrt{x y})} \\
&\leqslant \frac{2 x y(1-\sqrt{x y})}{2 \sqrt{x y}(1+\sqrt{x y})}=\frac{\sqrt{x y}(1-\sqrt{x y})}{1+\sqrt{x y}} .
\end{aligned}
$$

Let $t=\sqrt{x y}$. Then $t \in(0,1)$. So,
$$
\begin{aligned}
A &\leqslant \frac{t(1-t)}{1+t}=\frac{-t^{2}+t}{1+t} \\
&=\frac{-(t+1)^{2}+3(t+1)-2}{1+t} \\
&=3-\left(t+1+\frac{2}{t+1}\right) \\
&\leqslant 3-2 \sqrt{(t-1) \frac{2}{t+1}}=3-2 \sqrt{2} \text {. } \\
\end{aligned}
$$

When and only when $t=\sqrt{2}-1$, i.e., $\tan \frac{\alpha}{2}=\tan \frac{\beta}{2}=\sqrt{2}-1$, the equality holds.
Therefore, the maximum value of $A$ is $3-2 \sqrt{2}$.