1.已知sinα=(2√5)/5,求tan(α+π)+[sin(5/2π+α)/cos(5/2π-α)] 2.(1)若角α是第二象限角,化简

题目

1.已知sinα=(2√5)/5,求tan(α+π)+[sin(5/2π+α)/cos(5/2π-α)] 2.(1)若角α是第二象限角,化简
1.已知sinα=(2√5)/5,求tan(α+π)+[sin(5/2π+α)/cos(5/2π-α)]
2.(1)若角α是第二象限角,化简tan√[(1/sin^a)-1]
(2)化简：[√（1-2sin130°cos130°）]/[sin130°+√1-sin^2(130°)]

答案

1.
tan(α+π)+[sin(5/2π+α)/cos(5/2π-α)]
=tanα+[sin(α+π/2)/cos(-α+π/2)]
=tanα+[sin(α+π/2)/cos(-α+π/2)]
=tanα+[sin(π/2-α)/cos(π/2-α)]
=tanα+cosα/sinα
=(sinα/cosα)+(cosα/sinα)
=1/(sinαcosα)
cosα=±√(1-4/5)=√5/5,sinαcosα=±2/5
tan(α+π)+[sin(5/2π+α)/cos(5/2π-α)]=±5/2
2.
(1)角α是第二象限角,cosα<0
tanα√[(1/sin^α)-1]
=tanα·√[(1-sin^α)/sin^α]
=tanα·√(cos^α/sin^a)
=tanα·(-cosα/sinα)
=-1
(2)
[√（1-2sin130°cos130°）]/[sin130°+√1-sin^2(130°)]
=[√（1+2sin50°cos50°）]/[sin50°+√cos^2(130°)]
=[√（sin50°+cos50°）^2]/(sin50°+cos50°)
=（sin50°+cos50°）/(sin50°+cos50°)
=1