6. 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
$$
\frac{\sin A + \cos A \cdot \tan C}{\sin B + \cos B \cdot \tan C}
$$

the range of values is ( ).
(A) $(0,+\infty)$
(B) $\left(\frac{\sqrt{5}-1}{2},+\infty\right)$
(C) $\left(0, \frac{\sqrt{5}+1}{2}\right)$
(D) $\left(\frac{\sqrt{5}-1}{2}, \frac{\sqrt{5}+1}{2}\right)$
6. D.

Let the common ratio of $a$, $b$, and $c$ be $q$. Then $b=a q$, $c=a q^{2}$.
$$
\begin{array}{l}
\text { and } \frac{\sin A+\cos A \cdot \tan C}{\sin B+\cos B \cdot \tan C} \\
=\frac{\sin A \cdot \cos C+\cos A \cdot \sin C}{\sin B \cdot \cos C+\cos B \cdot \sin C} \\
=\frac{\sin (A+C)}{\sin (B+C)}=\frac{\sin (\pi-B)}{\sin (\pi-A)} \\
=\frac{\sin B}{\sin A}=\frac{b}{a}=q .
\end{array}
$$

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 must be that $a+b>c$ and $b+c>a$, i.e.,
$$
\left\{\begin{array}{l}
q^{2}-q-1<0 .
\end{array}\right.
$$

Solving this, we get $\frac{\sqrt{5}-1}{2}<q<\frac{\sqrt{5}+1}{2}$.