题目

（求步骤）化简：[sin(-α-3/2π)cos(3/2π-α)·tan2(π-α)]/[cos（π/2-α）·sin（π/2+α）]

提问时间：2020-07-16

答案

[sin(-α-3/2π)cos(3/2π-α)·tan2(π-α)]/[cos（π/2-α）·sin（π/2+α）]
=[cosa(-sina)(-tan2a)]/[sina(-cosa)]
=-tan2a