$\begin{array}{l} \left( {\dfrac{1}{{\cos 2x}} + 1} \right)\tan x\\ = \dfrac{{\cos 2x + 1}}{{\cos 2x}}.\tan x\\ = \dfrac{{2{{\cos }^2}x - 1 + 1}}{{\cos 2x}}.\tan x\\ = \dfrac{{2{{\cos }^2}x}}{{\cos 2x}}.\dfrac{{\sin x}}{{\cos x}} = \dfrac{{2\sin x.\cos x}}{{\cos 2x}} = \dfrac{{\sin 2x}}{{\cos 2x}} = \tan 2x \to A \end{array}$