Воспользуемся основным тригонометрическим тождеством:
\({\sin ^2}\alpha + {\cos ^2}\alpha = 1\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,{\sin ^2}\alpha = 1-\dfrac{{{m^2}}}{{{m^2} + {n^2}}}\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,{\sin ^2}\alpha = \dfrac{{{n^2}}}{{{m^2} + {n^2}}}\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\)
\( \Leftrightarrow \,\,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{\sin \alpha = \dfrac{{\left| n \right|}}{{\sqrt {{m^2} + {n^2}} }},\,\,\,}\\{\sin \alpha = -\dfrac{{\left| n \right|}}{{\sqrt {{m^2} + {n^2}} }}.}\end{array}} \right.\)
Так как \({270^ \circ } < \alpha < {360^ \circ }\) (IV четверть), то \(\sin \alpha < 0\), то есть \(\sin \alpha = -\dfrac{{\left| n \right|}}{{\sqrt {{m^2} + {n^2}} }}\).
\({\rm{tg}}\alpha + {\rm{ctg}}\alpha = \dfrac{{\sin \alpha }}{{\cos \alpha }} + \dfrac{{\cos \alpha }}{{\sin \alpha }} = \dfrac{{{{\sin }^2}\alpha + {{\cos }^2}\alpha }}{{\cos \alpha \sin \alpha }} = \dfrac{1}{{-\dfrac{m}{{\sqrt {{m^2} + {n^2}} }} \cdot \dfrac{{\left| n \right|}}{{\sqrt {{m^2} + {n^2}} }}}} = -\dfrac{{{m^2} + {n^2}}}{{m\left| n \right|}}.\)
Ответ: \(-\dfrac{{{m^2} + {n^2}}}{{m\left| n \right|}}.\)