2. Given $\alpha, \beta, \gamma$ are three unequal acute angles. If $\tan \alpha=\frac{\sin \beta \cdot \sin \gamma}{\cos \beta-\cos \gamma}$, then $\tan \beta$ equals ( ).
(A) $\frac{\sin \gamma \cdot \sin \alpha}{\cos \gamma-\cos \alpha}$
(B) $\frac{\sin \gamma \cdot \sin \alpha}{\cos \gamma+\cos \alpha}$
(C) $\frac{\cos \gamma-\cos \alpha}{\sin \gamma \cdot \sin \alpha}$
(D) $\frac{\cos \gamma+\cos \alpha}{\sin \gamma \cdot \sin \alpha}$
2. B.

From the given, we know that $\tan \alpha>0$,
$\sin \beta \cdot \sin \gamma>0$, then
$$
\cos \beta-\cos \gamma>0.
$$

Therefore, we can construct a right-angled $\triangle ABC$ as shown in Figure 4, such that $\angle C$
$$
\begin{array}{c}
=90^{\circ}, \angle A=\alpha, \\
AC=\cos \beta-\cos \gamma, \\
BC=\sin \beta \cdot \sin \gamma.
\end{array}
$$

Then $AB=\sqrt{(\cos \beta-\cos \gamma)^{2}+\sin ^{2} \beta \cdot \sin ^{2} \gamma}$
$$
\begin{array}{l}
=\sqrt{(\cos \beta-\cos \gamma)^{2}+\left(1-\cos ^{2} \beta\right)\left(1-\cos ^{2} \gamma\right)} \\
=\sqrt{(1-\cos \beta \cdot \cos \gamma)^{2}} \\
=1-\cos \beta \cdot \cos \gamma.
\end{array}
$$

Thus, $\sin \alpha=\frac{BC}{AB}=\frac{\sin \beta \cdot \sin \gamma}{1-\cos \beta \cdot \cos \gamma}$,
$$
\begin{array}{l}
\cos \alpha=\frac{AC}{AB}=\frac{\cos \beta-\cos \gamma}{1-\cos \beta \cdot \cos \gamma}. \\
\text { Therefore, } \frac{\sin \gamma \cdot \sin \alpha}{\cos \gamma+\cos \alpha}=\frac{\sin \gamma \cdot \frac{\sin \beta \cdot \sin \gamma}{1-\cos \beta \cdot \cos \gamma}}{\cos \gamma+\frac{\cos \beta-\cos \gamma}{1-\cos \beta \cdot \cos \gamma}} \\
=\frac{\sin \beta \cdot \sin ^{2} \gamma}{\cos \beta\left(1-\cos ^{2} \gamma\right)}=\tan \beta.
\end{array}
$$