2. If the three lateral edges of the tetrahedron $S-ABC$ are pairwise perpendicular, and $O$ is a point within the base $\triangle ABC$, then
$$
\mu=\tan \angle O S A \cdot \tan \angle O S B \cdot \tan \angle O S C
$$

the range of values for $\mu$ is $\qquad$
2. $\mu \geqslant 2 \sqrt{2}$.

From $\cos ^{2} \angle O S A+\cos ^{2} \angle O S B+\cos ^{2} \angle O S C=1$
$$
\begin{array}{l}
\Rightarrow \sin ^{2} \angle O S C=\cos ^{2} \angle O S A+\cos ^{2} \angle O S B \\
\quad \geqslant 2 \cos \angle O S A \cdot \cos \angle O S B .
\end{array}
$$

Similarly, we can obtain the other two inequalities.
Therefore, $\mu \geqslant 2 \sqrt{2}$.