[sin(2π+α)·tan(-α)·cos(-α)]÷[sin(-α)·cos(2π+α)·tan(2π+α)化简
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[sin(2π+α)·tan(-α)·cos(-α)]÷[sin(-α)·cos(2π+α)·tan(2π+α)化简
[sin(2π+α)·tan(-α)·cos(-α)]÷[sin(-α)·cos(2π+α)·tan(2π+α)化简
[sin(2π+α)·tan(-α)·cos(-α)]÷[sin(-α)·cos(2π+α)·tan(2π+α)化简
[sin(2π+α)·tan(-α)·cos(-α)]÷[sin(-α)·cos(2π+α)·tan(2π+α)]
=[sin(α)·sin(-α)]÷[sin(-α)·sin(2π+α)]
=[sin(α)·sin(-α)]÷[sin(-α)·sin(α)]
=1
求证(tanα·sinα)/(tanα-sinα)=(tanα+sinα)/(tanα·sinα)
化简:(sin^2α-tan^2α)/(sin^2α·tan^2α)
[sin(2π+α)·tan(-α)·cos(-α)]÷[sin(-α)·cos(2π+α)·tan(2π+α)化简
若α∈(-π/2+2kπ,2kπ)(k∈Z),则sinα,cosα,tanα的大小关系是A.tanα>sinα>cosαB.tanα>cosα>sinαC.tanα<sinα<cosαD.tanα<cosα<sinα
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cosα·tanα-Sinα
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sin²(-α)·cos(π+α)/tan(2∏+α)tan(∏+α)cos³(-∏-α)
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