\(tanx=\dfrac{4}{3}\)
\(\Rightarrow cotx=\dfrac{1}{tanx}=\dfrac{1}{\dfrac{4}{3}}=\dfrac{3}{4}\)
\(1+tan^2x=\dfrac{1}{cos^2x}\)
\(\Rightarrow cos^2x=\dfrac{1}{1+tan^2x}\)
\(=\dfrac{1}{1+\left(\dfrac{4}{3}\right)^2}=\dfrac{1}{1+\dfrac{16}{9}}=\dfrac{1}{\dfrac{25}{9}}=\dfrac{9}{25}\)
\(\Rightarrow cosx=\dfrac{3}{5}\)
\(sin^2x+cos^2x=1\)
\(\Rightarrow sin^2x=1-cos^2x=1-\left(\dfrac{3}{5}\right)^2=1-\dfrac{9}{25}=\dfrac{16}{25}\)
\(\Rightarrow sinx=\dfrac{4}{5}\)
Có \(tan.\alpha=\dfrac{4}{3}\)
Mà \(tan.\alpha.cot.\alpha=1\)
\(\Rightarrow cot.\alpha=1:\dfrac{4}{3}=\dfrac{3}{4}\)
Lại có \(sin^2\alpha+cos^2\alpha=1\\ \Leftrightarrow sin^2\alpha=1-cos^2\alpha\\ \Leftrightarrow sin\alpha=\sqrt{1-cos^2\alpha}\)
Vì \(tan.\alpha=\dfrac{sin.\alpha}{cos.\alpha}\)
\(\Leftrightarrow\dfrac{4}{3}=\dfrac{\sqrt{1-cos^2\alpha}}{cos.\alpha}\)
\(\Leftrightarrow\dfrac{4}{3}=\dfrac{1-cos^2\alpha}{cos^2\alpha}\\ \Leftrightarrow4.cos^2\alpha=3.\left(1-cos^2\alpha\right)\\ \Leftrightarrow4.cos^2\alpha=3-3cos^2\alpha\\ \Leftrightarrow cos.\alpha=\dfrac{\sqrt{21}}{7}\)
\(\Rightarrow sin.\alpha=\sqrt{1-\left(\dfrac{\sqrt{21}}{7}\right)^2}=\dfrac{4}{7}\)
\(tan=\dfrac{4}{3}\)
\(1+tan^2=\dfrac{1}{cos^2}\Leftrightarrow1+\left(\dfrac{4}{3}\right)^2=\dfrac{1}{cos^2}\Leftrightarrow cos^2=\dfrac{9}{25}\Leftrightarrow cos=\dfrac{3}{5}\)
\(sin=\sqrt{1-cos^2}=\sqrt{1-\left(\dfrac{3}{5}\right)^2}=\dfrac{4}{5}\)
\(cot=\dfrac{1}{tan}=1:\dfrac{4}{3}=\dfrac{3}{4}\)
