\(cotx=\dfrac{1}{tanx}=-\dfrac{1}{2}\)
\(1+tan^2x=\dfrac{1}{cos^2x}\\ \Leftrightarrow1+\left(-2\right)^2=\dfrac{1}{cos^2x}\\ \Leftrightarrow cosx=\dfrac{\sqrt{5}}{5}\left(-\dfrac{\pi}{2}< x< \dfrac{\pi}{2}\right)\)
\(tanx=\dfrac{sinx}{cosx}\\ \Leftrightarrow sinx=tanx.cosx=\left(-2\right).\dfrac{\sqrt{5}}{5}=-\dfrac{2\sqrt{5}}{5}\)