Ta có: \({\sin ^2}\alpha + {\cos ^2}\alpha = 1\) \( \Rightarrow \dfrac{1}{{16}} + {\cos ^2}\alpha = 1\) \( \Rightarrow {\cos ^2}\alpha = \dfrac{{15}}{{16}}\).
Mà \(\alpha \in \left( {{{90}^0};{{180}^0}} \right)\) nên \(\cos \alpha < 0\).
Vậy \(\cos \alpha = - \dfrac{{\sqrt {15} }}{4}\).
Suy ra \(\tan \alpha = \dfrac{{\sin \alpha }}{{\cos \alpha }}\)\( = \dfrac{1}{4}:\left( { - \dfrac{{\sqrt {15} }}{4}} \right) = - \dfrac{{\sqrt {15} }}{{15}}\).
