\(1+\cot^2a=\dfrac{1}{\sin^2a}=1+\dfrac{1}{4}=\dfrac{5}{4}\)
\(\Leftrightarrow\sin^2a=\dfrac{4}{5}\)
hay \(\sin a=-\dfrac{2\sqrt{5}}{5}\left(\Pi< a< \dfrac{3\Pi}{2}\right)\)
=>\(\cos a=-\dfrac{\sqrt{5}}{5}\)
\(\sin^2a\cdot\cos a=\dfrac{4}{5}\cdot\dfrac{-\sqrt{5}}{5}=\dfrac{-4\sqrt{5}}{25}\)