Question

j急！！！1.已知sinα=(2√5)/5,求tan(α+π)+[sin(5/2π+α)/cos(5/2π-α)]

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°)]

Answer 1

1. sinα=(2√5)/5 cossα=±(3√5)/5
tan(α+π)+[sin(5/2π+α)/cos(5/2π-α)]=tanα+[sin(1/2π+α)/cos(1/2π-α)]
=tanα+[sin(1/2π-α)/cos(1/2π-α)]=tanα+(cosα)/(sinα)=tanα+cotα
=1/(sinαcosα)=±5/6
2.
(1)tanα√[(1/sin^2a)-1]=tanα√[(1-sin^2a)/sin^2a]=tanα|cotα|
∵角α是第二象限角 ∴ cotα<0 ∴|cotα|=-cotα
∴原式+-1
(2)[√（1-2sin130°cos130°）]/[sin130°+√1-sin^2(130°)]
=[√（sin130°-cos130°）^2]/[sin130°+√(cos130°)^2]
=(sin130°-cos130°)/(sin130°-cos130°)
=1