2 2arc cos $\frac{3}{5}$ equals
A. $\arcsin \frac{24}{25}$
B. $\pi-\arctan \frac{24}{7}$
C. $\operatorname{arccot} \frac{7}{24}$
D. $-\arccos \frac{7}{25}$
2 A.
$$
\cos \left(2 \arccos \frac{3}{5}\right)=2\left(\frac{3}{5}\right)^{2}-1=-\frac{7}{25} .
$$

Since $0<\arccos \frac{3}{5}<\frac{\pi}{2}$, then $0<2 \arccos \frac{3}{5}<\pi$.
$$
\sin \left(2 \arccos \frac{3}{5}\right)=\sqrt{1-\left(\frac{7}{25}\right)^{2}}=\frac{24}{25}
$$

Therefore,
$$
2 \arccos \frac{3}{5}=\arcsin \frac{24}{25} \text {. }
$$