Ta có: \(\sin^2\alpha+\cos^2\alpha=1\)
\(\Leftrightarrow\cos^2\alpha=1-\sin^2\alpha=1-\left(\frac{1}{5}\right)^2=\frac{24}{25}\)
\(\Leftrightarrow\cos\alpha=\frac{2\sqrt{6}}{5}\)
\(\tan\alpha=\frac{\sin\alpha}{\cos\alpha}=\frac{\frac{1}{5}}{\frac{2\sqrt{6}}{5}}=\frac{\sqrt{6}}{12}\)
\(\tan\alpha.\cot\alpha=1\)
\(\Leftrightarrow\cot\alpha=\frac{1}{\tan\alpha}=\frac{1}{\frac{\sqrt{6}}{12}}=2\sqrt{6}\)
Chúc bạn hok tốt!!! nguyenthitonga