Sửa đề: \(\frac{\pi}{2}
=>\(\sin a>0;\cos a<0;\tan a<0\)
Ta có: \(1+\cot^2a=\frac{1}{\sin^2a}\)
=>\(\frac{1}{\sin^2a}=1+\left(-3\right)^2=1+9=10\)
=>\(sin^2a=\frac{1}{10}\)
mà sin a>0
nên \(\sin a=\sqrt{\frac{1}{10}}=\frac{\sqrt{10}}{10}\)
Ta có: \(\tan a\cdot\cot a=1\)
=>\(\tan a=\frac{1}{\cot a}=\frac{1}{-3}=\frac{-1}{3}\)
=>\(\frac{\sin a}{\cos a}=\frac{-1}{3}\)
=>\(cosa=-3\cdot\sin a=-3\cdot\frac{\sqrt{10}}{10}=\frac{-3\sqrt{10}}{10}\)