`a)` Ta có :
$sin^2\alpha+cos^2\alpha=1$
`=>`$cos^2\alpha=1-sin^2\alpha$
`=>`$cos^2\alpha=1-(0,8)^2=1-0,64=0,36$
`=>`$cos\alpha=±0,6$
Ta cũng có :
$tan\alpha=$`(sin\alpha)/cos\alpha `
`=>`$tan\alpha=$`(0,8)/(±0,6)=±4/3 `
$cot\alpha=$`cos\alpha/(sin\alpha)`
$cot\alpha=$`(±0,6)/(0,8)=±3/4`
Vậy $sin\alpha=0,8;cos\alpha=0,6;tan\alpha=$`±4/3`$;cot\alpha=$`±3/4`
`b)` Ta có :
$tan\alpha·cot\alpha=1$
`=>`$cot\alpha=$`1/tan\alpha`
`=>`$cot\alpha=$`1/3`
Ta cũng có :
$1+tan^2 a=\dfrac{1}{cos^2 a}$
`<=>` $1+9=\dfrac{1}{cos^2 a}$
`<=>`$\cos^2a=\dfrac1{10}$
`=>`$cos a=\pm$`\sqrt{10}/10`
$tan\alpha=$`(sin\alpha)/cos\alpha `
`=>`$sin\alpha$`=tan\alpha·cos\alpha`
`=>`$sin\alpha$`=3·(\pm\sqrt{10}/10)`
`=>sin\alpha=`$\pm$`(\3sqrt{10})/10`
Vậy $sin\alpha=\pm$`(\3sqrt{10})/10`;$cos\alpha=\pm$`\sqrt{10}/10`;$tan\alpha=3$$;cot\alpha=$`1/3`
