ta có
\(sin^2x+cos^2x=1\Leftrightarrow sin^2x=1-cos^2x=1-0.6^2=0.64\)
TH1.\(sinx=\sqrt{0.64}=0.8\Rightarrow\hept{\begin{cases}tanx=\frac{sinx}{cosx}=\frac{0.8}{0.6}=\frac{4}{3}\\cotx=\frac{1}{tanx}=\frac{3}{4}\end{cases}}\)
TH2.\(sinx=-\sqrt{0.64}=-0.8\Rightarrow\hept{\begin{cases}tanx=\frac{sinx}{cosx}=\frac{-0.8}{0.6}=-\frac{4}{3}\\cotx=\frac{1}{tanx}=-\frac{3}{4}\end{cases}}\)