6. As shown in Figure 1, in the octahedron $P A B C D Q$ inscribed in sphere $O$, vertices $P$ and $Q$ are on opposite sides of plane $A B C D$, and the pyramids $P-A B C D$ and $Q-A B C D$ are both regular square pyramids. Let the size of the dihedral angle $P-A B-Q$ be $\theta$. Then the range of $\tan \theta$ is
6. $(-\infty,-2 \sqrt{2}]$.

Let the dihedral angle $P-A B-C$ be $\alpha$, and the dihedral angle $Q-A B-C$ be $\beta$. The distance from the center of the sphere to the plane $A B C D$ is $d$, and the radius of the sphere is 1. Then,
$$
\tan \alpha=\frac{\sqrt{2}(1-d)}{\sqrt{1-d^{2}}}, \tan \beta=\frac{\sqrt{2}(1+d)}{\sqrt{1-d^{2}}},
$$

where, $d \in[0,1)$.
$$
\begin{array}{l}
\text { Hence } \tan \theta=\tan (\alpha+\beta) \\
=\frac{\frac{\sqrt{2}(1-d)}{\sqrt{1-d^{2}}}+\frac{\sqrt{2}(1+d)}{\sqrt{1-d^{2}}}}{1-\frac{\sqrt{2}(1-d)}{\sqrt{1-d^{2}}} \cdot \frac{\sqrt{2}(1+d)}{\sqrt{1-d^{2}}}} \\
=\frac{-2 \sqrt{2}}{\sqrt{1-d^{2}}} \in(-\infty,-2 \sqrt{2}] .
\end{array}
$$