詳解：\(f(x) = \sec x{\rm{ = }}\frac{{\rm{1}}}{{\cos x}}\quad  \Rightarrow \quad f'(x) = \frac{{{\rm{0}} \cdot \cos x - 1 \cdot (\cos x)'}}{{{{\left( {\cos x} \right)}^2}}} = \frac{{{\rm{0}} \cdot \cos x - 1 \cdot (\cos x)'}}{{{{\cos }^2}x}}\)m\)

\( f'(x) = \frac{{\sin x}}{{{{\cos }^2}x}} = \frac{1}{{\cos x}} \cdot \frac{{\sin x}}{{\cos x}} = \sec x\tan x\)

故選(A)