[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(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