Cho \(\sin \alpha = \frac{3}{5},\left( {90^\circ < \alpha < 180^\circ } \right)\).

a) \(\cos \alpha > 0\).

b) \({\cos ^2}\alpha = \frac{{16}}{{25}}\).

c) \[\tan \left( {180^\circ - \alpha } \right) = - \frac{3}{4}\].

d) \[A = \frac{{\tan \alpha - \cot \left( {180^\circ - \alpha } \right)}}{{\sin \left( {90^\circ - \alpha } \right)}} = \frac{{125}}{{48}}\].

a) Sai. Vì \(90^\circ < \alpha < 180^\circ \) nên \(\cos \alpha < 0\).

b) Đúng. Ta có \({\sin ^2}\alpha + {\cos ^2}\alpha = 1 \Rightarrow {\cos ^2}\alpha = 1 - {\sin ^2}\alpha = 1 - {\left( {\frac{3}{5}} \right)^2} = \frac{{16}}{{25}}\).

Do đó \[\cos \alpha = - \sqrt {\frac{{16}}{{25}}} = - \frac{4}{5}\].

c) Sai. Ta có \[\tan \alpha = \frac{{\sin \alpha }}{{\cos \alpha }} = - \frac{3}{4} \Rightarrow \,\tan \left( {180^\circ - \alpha } \right) = - \tan \alpha = \frac{3}{4}\].

d) Đúng. Ta có \(\cot \alpha = \frac{1}{{\tan \alpha }} = - \frac{4}{3}\).

Vậy \[A = \frac{{\tan \alpha - \cot \left( {180^\circ - \alpha } \right)}}{{\sin \left( {90^\circ - \alpha } \right)}} = \frac{{\tan \alpha + \cot \alpha }}{{\cos \alpha }} = \frac{{\left( { - \frac{3}{4}} \right) + \left( { - \frac{4}{3}} \right)}}{{ - \frac{4}{5}}} = \frac{{125}}{{48}}\].