[sin(2π+α）·tan(-α)·cos(-α)]÷[sin(-α)·cos(2π+α）·tan(2π+α）化简_数学解答

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

人气：206 ℃　时间：2020-10-01 04:59:43

解答

[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