2.2.45 ** In $\triangle A B C$, if $\frac{\cos A}{\sin B}+\frac{\cos B}{\sin A}=2$, and the perimeter of $\triangle A B C$ is 12. Find the maximum possible value of its area.
Given that $\sin \Lambda(\cos A-\sin B)+\sin B(\cos B-\sin A)=0$, we have
$$
\begin{aligned}
& \sin A\left[\sin \left(90^{\circ}-A\right)-\sin B\right]+\sin B\left[\sin \left(90^{\circ}-B\right)-\sin A\right] \\
= & 2 \sin \frac{90^{\circ}-A-B}{2}\left[\sin A \cdot \cos \left(45^{\circ}-\frac{A-B}{2}\right)+\sin B \cdot \cos \left(45^{\circ}+\frac{A-B}{2}\right)\right] \\
= & 2 \sin \frac{90^{\circ}-A-B}{2} \times \frac{\sqrt{2}}{2} \times\left[\cos \frac{A-B}{2} \cdot(\sin A+\sin B)+\sin \frac{A-B}{2} \cdot(\sin A-\sin B)\right] \\
= & 0 .
\end{aligned}
$$

And
$$
\begin{aligned}
& \cos \frac{A-B}{2} \cdot(\sin A+\sin B)+\sin \frac{A-B}{2} \cdot(\sin A-\sin B) \\
= & 2 \cos ^{2} \frac{A-B}{2} \sin \frac{A+B}{2}+2 \cos \frac{A+B}{2} \cdot \sin ^{2} \frac{A-B}{2}>0,
\end{aligned}
$$

Thus, $\sin \frac{90^{\circ}-A-B}{2}=0$. Therefore, $90^{\circ}-\angle A-\angle B=0, \angle A+\angle B=90^{\circ}$, which means $\triangle A B C$ is a right triangle.

Let the sides opposite to $A$, $B$, and $C$ be $a$, $b$, and $c$ respectively. According to the problem, we have $a+b+\sqrt{a^{2}+b^{2}}=12$. Since $12=a+b+\sqrt{a^{2}+b^{2}} \geqslant 2 \sqrt{a b}+\sqrt{2 a b}$, it follows that $a b \leqslant 36(2-\sqrt{2})^{2}$. Therefore, $S=\frac{1}{2} a b \leqslant 18(2-\sqrt{2})^{2}=36(3-2 \sqrt{2})$, so $S_{\text {max }}=36(3-2 \sqrt{2})$.