Xét ΔABC vuông tại A có
\(sinC=cosB\)
=>\(sinC=\dfrac{3}{5}\)
\(sin^2C+cos^2C=1\)
=>\(cos^2C=1-\left(\dfrac{3}{5}\right)^2=\dfrac{16}{25}\)
=>\(cosC=\dfrac{4}{5}\)
\(tanC=\dfrac{sinC}{cosC}=\dfrac{3}{5}:\dfrac{4}{5}=\dfrac{3}{4}\)
\(cotC=\dfrac{cosC}{sinC}=\dfrac{4}{5}:\dfrac{3}{5}=\dfrac{4}{3}\)