[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

题目

[sin(2π+α）·tan(-α)·cos(-α)]÷[sin(-α)·cos(2π+α）·tan(2π+α）化简

答案

[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

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