`a)` `sin α=4/ 5`
Ta có:
`\qquad sin^2α+cos^α=1`
`=>cos^2α=1-sin^2α=1-(4/ 5)^2=9/{25}`
`=>`$\left[\begin{array}{l}cosα=\dfrac{3}{5}\\cosα=\dfrac{-3}{5}\end{array}\right.$
Vì `π/ 2<α<π=>-1<cosα<0`
`=>cosα=-3/ 5`
$\\$
`cos2α=2cos^2α-1=2. 9/{25}-1={-7}/{25}`
Vậy `cosα=-3/ 5; cos2α=-7/{25}`
$\\$
`b)` Áp dụng:
`sin(a+b)=sinacosb+sinbcosa`
`cos(a+b)=cosacosb-sinasinb`
`tan(π-α)=-tanα`
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Ta có:
`\qquad tanA. cotB+1`
`={sinA}/{cosA}.{cosB}/{sinB}+1`
`={sinAcosB+sinBcosA}/{cosAsinB}`
`={sin(A+B)}/{cosAsinB}`
$\\$
`\qquad tanA-cotB`
`={sinA}/{cosA}-{cosB}/{sinB}`
`={sinA sinB-cosAcosB}/{cosAsinB}`
`={-(cosAcosB-sinAsinB)}/{cosAsinB}`
`=-{cos(A+B)}/{cosAsinB}`
$\\$
`VT= {tanA.cotB +1}/{tanA-cotB}`
`={{sin(A+B)}/{cosAsinB}}/{{-cos(A+B)}/{cosAsinB}}`
`={sin(A+B)}/{-cos(A+B)}`
`=-tan(A+B)`
`=tan(π-(A+B))=tanC=VP`
Vậy: `{tanA.cotB +1}/{tanA-cotB}=tanC`
