$\begin{array}{l} \dfrac{\pi }{2} < \alpha < \pi \to \sin \alpha > 0,\cos \alpha < 0\\ \sin \alpha = \sqrt {1 - {{\cos }^2}\alpha } = \sqrt {1 - {{\left( { - \dfrac{{12}}{{13}}} \right)}^2}} = \sqrt {\dfrac{{25}}{{169}}} = \dfrac{5}{{13}}\\ \tan \alpha = \dfrac{{\sin \alpha }}{{\cos \alpha }} = \dfrac{5}{{13}}:\dfrac{{ - 12}}{{13}} = - \dfrac{5}{{12}}\\ \to D \end{array}$