1. 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.
1. From the given, $\sin A \cdot \cos A + \sin B \cdot \cos B = 2 \sin A \cdot \sin B$, which means
$$
\sin A(\cos A - \sin B) + \sin B(\cos B - \sin A) = 0
$$

Thus, $\sin A\left[\sin \left(90^{\circ} - A\right) - \sin B\right] + \sin B\left[\sin \left(90^{\circ} - B\right) - \sin A\right]$
$$
\begin{array}{l}
= 2 \sin \frac{90^{\circ} - A - B}{2}\left[\sin A \cdot \cos \left(45^{\circ} - \frac{A - B}{2}\right) + \sin B \cdot \cos \left(45^{\circ} + \frac{A - B}{2}\right)\right] \\
= 2 \sin \frac{90^{\circ} - A - B}{2} \cdot \frac{\sqrt{2}}{2} \cdot \left[\cos \frac{A - B}{2} \cdot (\sin A + \sin B) + \sin \frac{A - B}{2} (\sin A - \sin B)\right] \\
= 0
\end{array}
$$

Also, $\cos \frac{A - B}{2} \cdot (\sin A + \sin B) + \sin \frac{A - B}{2} (\sin A - \sin B) = 2 \cos^2 \frac{A - B}{2} \sin \frac{A + B}{2} + 2 \cos \frac{A + B}{2} \sin^2 \frac{A - B}{2} > 0$, so $\sin \frac{90^{\circ} - A - B}{2} = 0$. Therefore, $90^{\circ} - A - B = 0$, which means $A + B = 90^{\circ}$. Hence, $\triangle ABC$ is a right triangle. Let the sides opposite to $A$, $B$, and $C$ be $a$, $b$, and $c$ respectively. According to the problem, $a + b + \sqrt{a^2 + b^2} = 12$. Since $a + b \geq 2 \sqrt{ab}$ and $a^2 + b^2 \geq 2ab$, we have $12 = a + b + \sqrt{a^2 + b^2} \geq 2 \sqrt{ab} + \sqrt{2ab}$, which implies $ab \leq 36(2 - \sqrt{2})^2$. Therefore, $S = \frac{1}{2} ab \leq 18(2 - \sqrt{2})^2 = 36(3 - 2\sqrt{2})$, so $S_{\text{max}} = 36(3 - 2\sqrt{2})$.