$\begin{array}{l} t = \tan a = - 2 \to \sin 2a = \dfrac{{2t}}{{1 + {t^2}}},\cos 2a = \dfrac{{1 - {t^2}}}{{1 + {t^2}}}\\ \to A = 2\cos 2a - \sin 2a = \dfrac{{2\left( {1 - {t^2}} \right)}}{{1 + {t^2}}} - \dfrac{{2t}}{{1 + {t^2}}}\\ \to A = \dfrac{{2 - 2{t^2} - 2t}}{{1 + {t^2}}} = \dfrac{{2 - 2{{\left( { - 2} \right)}^2} + 2\left( { - 2} \right)}}{{1 + {{( - 2)}^2}}} = -\dfrac{2}{5} \end{array}$
