Ta có:
\(sin^2x+cos^2x=1\)
\(sin^2x+\left(\dfrac{3}{5}\right)^2=1\)
\(sin^2x=\dfrac{16}{25}\)
Vì \(180^o< x< 270^o\) \(\Rightarrow sinx< 0\)
\(\Rightarrow sinx=-\dfrac{4}{5}\)
\(tanx=\dfrac{sinx}{cosx}=\dfrac{-\dfrac{4}{5}}{\dfrac{3}{5}}=-\dfrac{4}{3}\)
\(cotx=\dfrac{cosx}{sinx}=\dfrac{\dfrac{3}{5}}{-\dfrac{4}{5}}=-\dfrac{3}{4}\)