4. Let the internal angles $\angle A, \angle B, \angle C$ of $\triangle ABC$ correspond to the sides $a, b, c$ which form a geometric sequence. Then the range of $\frac{\sin A \cdot \cot C+\cos A}{\sin B \cdot \cot C+\cos B}$ is $\qquad$
4. $\left(\frac{\sqrt{5}-1}{2}, \frac{\sqrt{5}+1}{2}\right)$.

Let the common ratio of $a$, $b$, and $c$ be $q$. Then
$$
b=a q, c=a q^{2} \text {. }
$$

Notice that,
$$
\begin{array}{l}
\frac{\sin A \cdot \cot C+\cos A}{\sin B \cdot \cot C+\cos B} \\
=\frac{\sin A \cdot \cos C+\cos A \cdot \sin C}{\sin B \cdot \cos C+\cos B \cdot \sin C} \\
=\frac{\sin (A+C)}{\sin (B+C)}=\frac{\sin (\pi-B)}{\sin (\pi-A)} \\
=\frac{\sin B}{\sin A}=\frac{b}{a} \\
=q .
\end{array}
$$

Therefore, we only need to find the range of $q$.
Since $a$, $b$, and $c$ form a geometric sequence, the largest side can only be $a$ or $c$. Thus, for $a$, $b$, and $c$ to form the sides of a triangle, it is necessary and sufficient that $a+b>c$ and $b+c>a$, i.e.,
$$
\begin{array}{l}
\left\{\begin{array}{l}
a+a q>a q^{2}, \\
a q+a q^{2}>a
\end{array}\right. \\
\Rightarrow\left\{\begin{array}{l}
q^{2}-q-1<0
\end{array}\right. \\
\Rightarrow\left\{\begin{array}{l}
\frac{1-\sqrt{5}}{2}<q<\frac{1+\sqrt{5}}{2}
\end{array}\right. \\
\Rightarrow \frac{\sqrt{5}-1}{2}<q<\frac{\sqrt{5}+1}{2} .
\end{array}
$$