Giúp mik vs ạ,mik đnag cần gấp ạ

$\begin{array}{l}
\dfrac{1}{{{{\cos }^2}x - {{\sin }^2}x}} + \dfrac{{2\tan x}}{{1 - {{\tan }^2}x}}\\
 = \dfrac{1}{{\cos 2x}} + \tan 2x\\
 = \dfrac{1}{{\cos 2x}} + \dfrac{{\sin 2x}}{{\cos 2x}}\\
 = \dfrac{{\sin 2x + 1}}{{\cos 2x}} = \dfrac{{\left( {\sin 2x + 1} \right)\cos 2x}}{{{{\cos }^2}2x}}\\
 = \dfrac{{\left( {\sin 2x + 1} \right)\cos 2x}}{{1 - {{\sin }^2}2x}}\\
 = \dfrac{{\left( {\sin 2x + 1} \right)\cos 2x}}{{\left( {1 - \sin 2x} \right)\left( {1 + \sin 2x} \right)}}\\
 = \dfrac{{\cos 2x}}{{1 - \sin 2x}} \Rightarrow a = 2,b = 1 \Rightarrow a + b = 3
\end{array}$