`\qquad tanα=1/ 2;cosα<0`
`=>cotα=1/{tanα}=1/{1/ 2}=2`
Ta có:
`\qquad tan^2α+1={sin^2α}/{cos^2α}+1={sin^2α+cos^2α}/{cos^2α}=1/{cos^2α}`
`=>(1/ 2)^2+1=1/ {cos^2α}`
`=>5/ 4 =1/{cos^2α}`
`=>cos^2α=4/ 5`
$⇒\left[\begin{array}{l}cosα=\dfrac{2}{\sqrt{5}}(loại)\\cosα=\dfrac{-2}{\sqrt{5}}\end{array}\right.$
`=>cosα={-2}/{\sqrt{5}}`
$\\$
`\qquad tanα={sinα}/{cosα}`
`=>sinα=tanα.cosα=1/ 2 . {-2}/{\sqrt{5}}={-1}/{\sqrt{5}}`
$\\$
Vậy: `sinα={-1}/{\sqrt{5}}; cosα={-2}/{\sqrt{5}}`
`\qquad tanα=1/ 2;cotα=2`
