Có: `tanA=1/3`
`=> cotA=1/(tanA)=3`
Lại có: `1+tan^2A=1/(cos^2A)`
`=> 1/(cos^2A)=1+(1/3)^2=10/9`
`=> cos^2A=9/10`
`=> cosA=+-(3\sqrt{10})/10`
Khi đó `sin^2A+cos^2A=1`
`<=> sin^2A=1-cos^2A=1-9/10=1/10`
`=> sinA=+-\sqrt{10}/10`
Vậy `sinA=+-\sqrt{10}/10;cosA=+-(3\sqrt{10})/10;cotA=3`