2. (2003 Hunan Province Competition Question) As shown in the figure, $S-ABC$ is a tetrahedron with three edges mutually perpendicular, and $O$ is a point within the base $ABC$. If $\angle OSA=\alpha, \angle OSB=\beta, \angle OSC=\gamma$, then the range of $\tan \alpha \cdot \tan \beta \cdot \tan \gamma$ is ( ).
A. $[2 \sqrt{2},+\infty]$
B. $(0,2 \sqrt{2})$
C. $[1,2 \sqrt{2}]$
D. $(1,2 \sqrt{2})$
2. Choose A. Reason: As shown in the figure, draw planes through $O$ parallel to $S A$, $S B$, and $S C$, intersecting $S A$, $S B$, and $S C$ at $D$, $E$, and $F$, respectively, thus forming a rectangular prism $M F S E-O N D P$ with $S O$ as the diagonal. From the problem, we know
$$
\begin{array}{l}
\cos ^{2} \alpha+\cos ^{2} \beta+\cos ^{2} \gamma=1 . \\
\text { Also, } \sin ^{2} \gamma=\cos ^{2} \alpha+\cos ^{2} \beta \geqslant 2 \cos \alpha \cdot \cos \beta, \\
\sin ^{2} \alpha=\cos ^{2} \beta+\cos ^{2} \gamma \geqslant 2 \cos \beta \cdot \cos \gamma, \\
\sin ^{2} \beta=\cos ^{2} \alpha+\cos ^{2} \gamma \geqslant 2 \cos \alpha \cdot \cos \gamma,
\end{array}
$$

From $\sin ^{2} \gamma=\cos ^{2} \alpha+\cos ^{2} \beta \geqslant 2 \cos \alpha \cdot \cos \beta$,

we get $\tan \alpha \cdot \tan \beta \cdot \tan \gamma \geqslant 2 \sqrt{2}$.