\(\sin \left( {\alpha -{{90}^ \circ }} \right) = \dfrac{{2\sqrt 6 }}{5}\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,-\cos \alpha = \dfrac{{2\sqrt 6 }}{5}\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\cos \alpha = -\dfrac{{2\sqrt 6 }}{5}.\)
Воспользуемся основным тригонометрическим тождеством:
\({\sin ^2}\alpha + {\cos ^2}\alpha = 1\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,{\left( {-\dfrac{{2\sqrt 6 }}{5}} \right)^2} + {\sin ^2}\alpha = 1\,\,\,\,\,\,\, \Leftrightarrow \)
\( \Leftrightarrow \,\,\,\,\,\,\,{\sin ^2}\alpha = \dfrac{1}{{25}}\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{\sin \alpha = \dfrac{1}{5},\,\,\,}\\{\sin \alpha = -\dfrac{1}{5}.}\end{array}} \right.\)
Так как \(\alpha \, \in \,\left( {{{90}^ \circ };{{180}^ \circ }} \right)\) (II четверть), то \(\sin \alpha > 0\), то есть \(\sin \alpha = \dfrac{1}{5}.\)
Тогда: \({\rm{tg}}\alpha = \dfrac{{\sin \alpha }}{{\cos \alpha }} = \dfrac{1}{5}:\left( {-\dfrac{{2\sqrt 6 }}{5}} \right) = -\dfrac{1}{{2\sqrt 6 }}.\)
\({\rm{tg}}2\alpha = \dfrac{{2{\rm{tg}}\alpha }}{{1-{\rm{t}}{{\rm{g}}^2}\alpha }} = \dfrac{{2 \cdot \left( {-\dfrac{1}{{2\sqrt 6 }}} \right)}}{{1-\dfrac{1}{{24}}}} = -\dfrac{1}{{\sqrt 6 }} \cdot \dfrac{{24}}{{23}} = -\dfrac{{4\sqrt 6 }}{{23}}.\)
Ответ: \(-\dfrac{{4\sqrt 6 }}{{23}}.\)