2. Let $\angle A$, $\angle B$, $\angle C$ all be acute angles. Given the following three propositions:
(1) $\tan \frac{A}{2}=\tan ^{3} \frac{C}{2}$, (2) $\tan C=2 \tan B$,
(3) $\angle A$, $\angle B$, $\angle C$ form an arithmetic sequence.
Then the number of true propositions derived from any two of (1), (2), (3) is ( ).
(A) 3
(B) 2
(C) 1
(D) 0
2.A.

Notice
$$
\begin{array}{l} 
\tan B=\frac{1}{2} \tan C=\frac{\tan \frac{C}{2}}{1-\tan ^{2} \frac{C}{2}} \\
=\frac{\tan \frac{C}{2}\left(1+\tan ^{2} \frac{C}{2}\right)}{1-\tan ^{4} \frac{C}{2}}=\frac{\tan \frac{C}{2}+\tan ^{3} \frac{C}{2}}{1-\tan ^{4} \frac{C}{2}} \\
=\frac{\tan \frac{C}{2}+\tan \frac{A}{2}}{1-\tan \frac{C}{2} \cdot \tan \frac{A}{2}}=\tan \left(\frac{A}{2}+\frac{C}{2}\right) . \\
\text { Also, } 0<\frac{\angle A}{2}+\frac{\angle C}{2}<\frac{\pi}{2}, 0<\angle B<\frac{\pi}{2}, \\
\angle B=\frac{\angle A}{2}+\frac{\angle C}{2},
\end{array}
$$

That is, $2 \angle B=\angle A+\angle C, \angle A, \angle B, \angle C$ form an arithmetic sequence, hence (1)(2) $\Rightarrow$ (3).
Similarly, (1)(3) $\Rightarrow$ (2).
$$
\begin{array}{l}
\text { Also, } \tan \left(\frac{A}{2}+\frac{C}{2}\right)=\frac{\tan \frac{C}{2}}{1-\tan ^{2} \frac{C}{2}}=\frac{1}{2} \tan C \\
=\tan B=\frac{\tan \frac{C}{2}+\tan \frac{A}{2}}{1-\tan \frac{C}{2} \cdot \tan \frac{A}{2}} \\
\Rightarrow \frac{\tan \frac{C}{2}}{1-\tan ^{2} \frac{C}{2}}=\frac{\tan \frac{C}{2}+\tan \frac{A}{2}}{1-\tan \frac{C}{2} \cdot \tan \frac{A}{2}} \\
\Rightarrow \tan \frac{A}{2}=\tan ^{3} \frac{C}{2},
\end{array}
$$

Therefore, (2)(3) $\Rightarrow$ (1).