(C) $\csc A$
Justification: Since $\sec^2 A - 1 = \tan^2 A$, we get $\dfrac{\sec A}{\sqrt{\tan^2 A}} = \dfrac{\sec A}{\tan A} = \dfrac{1/\cos A}{\sin A/\cos A} = \dfrac{1}{\sin A} = \csc A$.
Use the identity $\sec^2 A - \tan^2 A = 1 \Rightarrow \sec^2 A - 1 = \tan^2 A$. Substituting simplifies the expression to $\sec A / \tan A$, which on cancelling $\cos A$ gives $1/\sin A = \csc A$. Don't forget to take the positive square root (acute angle assumed).