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[sin(2π+α)·tan(-α)·cos(-α)]÷[sin(-α)·cos(2π+α)·tan(2π+α)化简
题目详情
[sin(2π+α)·tan(-α)·cos(-α)]÷[sin(-α)·cos(2π+α)·tan(2π+α)化简
▼优质解答
答案和解析
[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
=[sin(α)·sin(-α)]÷[sin(-α)·sin(2π+α)]
=[sin(α)·sin(-α)]÷[sin(-α)·sin(α)]
=1
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