6. Let the internal angles $A, B, C$ of $\triangle ABC$ correspond to the sides $a, b, c$ which form a geometric sequence. Then the range of $\frac{\sin A \cot C + \cos A}{\sin B \cot C + \cos B}$ is
A. $(0,+\infty)$
B. $\left(0, \frac{\sqrt{5}+1}{2}\right)$
C. $\left(\frac{\sqrt{5}-1}{2}, \frac{\sqrt{5}+1}{2}\right)$
D. $\left(\frac{\sqrt{5}-1}{2},+\infty\right)$
Let the common ratio of the geometric sequence $a, b, c$ be $q$, then $b=a q, c=a q^{2}$.
Thus, $\left\{\begin{array}{l}a+b>c, \\ b+c>a\end{array} \Rightarrow\left\{\begin{array}{l}q^{2}-q-1<0\end{array} \Rightarrow \frac{\sqrt{5}-1}{2}<q<\frac{\sqrt{5}+1}{2}\right.\right.$.
$\frac{\sin A \cot C+\cos A}{\sin B \cot C+\cos B}=\frac{\sin A \cos C+\cos A \sin C}{\sin B \cos C+\cos B \sin C}=\frac{\sin (A+C)}{\sin (B+C)}$
$=\frac{\sin B}{\sin A}=\frac{b}{a}=q \in\left(\frac{\sqrt{5}-1}{2}, \frac{\sqrt{5}+1}{2}\right)$. Therefore, the answer is $C$.