One. (15 points) As shown in Figure 1, $A$ is a fixed point between two parallel lines $l_{1}$ and $l_{2}$, and the distances from point $A$ to lines $l_{1}$ and $l_{2}$ are $A M=1, A N=\sqrt{3}$. Let the other two vertices $C$ and $B$ of $\triangle A B C$ move on $l_{1}$ and $l_{2}$ respectively, and satisfy $A B<A C, \frac{A B}{\cos B}=\frac{A C}{\cos C}$.
(1) Determine the shape of $\triangle A B C$ and prove the conclusion;
(2) Find the maximum value of $\frac{1}{A B}+\frac{\sqrt{3}}{A C}$.
(1) From $\frac{A B}{\cos B}=\frac{A C}{\cos C}$ and the Law of Sines, we get
$$
\frac{\sin C}{\cos B}=\frac{\sin B}{\cos C} \Rightarrow \sin 2 B=\sin 2 C \text {. }
$$

Since $A B<A C$, we have $\angle C<\angle B$. Therefore, $2 \angle B+2 \angle C=\pi \Rightarrow \angle B+\angle C=\frac{\pi}{2}$. Hence, $\triangle A B C$ is a right triangle, and $\angle A=\frac{\pi}{2}$.
(2) Let $\angle B A M=\theta\left(0<\theta<\frac{\pi}{2}\right)$. Then $\angle C A N=\frac{\pi}{2}-\theta$.
Thus, in the right $\triangle A M B$, $A B=\frac{1}{\cos \theta}$; in the right $\triangle A N C$, $A C=\frac{\sqrt{3}}{\sin \theta}$. Therefore, $\frac{1}{A B}+\frac{\sqrt{3}}{A C}=\cos \theta+\sin \theta=2 \sin \left(\theta+\frac{\pi}{4}\right)$. Hence, when $\theta=\frac{\pi}{4}$, $\frac{1}{A B}+\frac{\sqrt{3}}{A C}$ reaches its maximum value of $\sqrt{2}$.