4. In the tetrahedron $O A B C$, $O A \perp$ plane $A B C, A B \perp$ $A C$, point $P$ satisfies $O P=l O A+m O B+n O C$, where $l, m, n$ are positive numbers and $l+m+n=1$. If the line $O P$ is formed by points that are equidistant from the planes $O B C$, $O C A$, and $O A B$, then the cosine value of the dihedral angle $A-O B-C$ is (expressed in terms of $l, m, n$)
4. $\frac{n}{l}$.

From the conditions, we know that point $P$ is inside $\triangle A B C$ and the area ratios of $\triangle P B C$, $\triangle P C A$, and $\triangle P A B$ are $l: m: n$. Furthermore, from the composition of $O P$, we know that the area ratios of $\triangle O B C$, $\triangle O C A$, and $\triangle O A B$ are also $l: m: n$.

Therefore, the cosine value of the dihedral angle $A-O B-C$ is the area ratio of $\triangle O A B$ to $\triangle O B C$, which should be $\frac{n}{l}$.