Question

化简sin²(α+π)·cos(π+α)·cot(-α-2π)/tan(π+α)·cos²(-α-π)求答案

Answer 1

解sin²(α+π)·cos(π+α)·cot(-α-2π)/tan(π+α)·cos²(-α-π)
=sin²(α)·（-cos(α)）·（-cot(α+2π)）/（tan(α)）·cos²(α)
=sin²(α)·cos(α)·cot(α+2π)/（tan(α)）·cos²(α)
=sin²(α)·cos(α)·cot(α)/（tan(α)）·cos²(α)
=sin²(α)·cos(α)/（tan²(α）·cos²(α)
=sin²(α)·cos(α)/tan²(α)·cos²(α)
=sin²(α)·cos(α)/[sin²(α)/cos²a·cos²(α)]
=sin²(α)·cos(α)/[sin²(α)]
=cosa

追问

您好，答案上的得数是-1，您确定是对的吗？

追答

确认

Answer 2

[sin²(α+π)·cos(π+α)·cot(-α-2π)]/[tan(π+α)·cos²(-α-π)]
=[sin²α(-cosα)(-cotα)]/[tanαcos²α]
=(sin²αcosα*cosα/sinα)/(sinα/cosα*cos²α)
=cosα

追问

您好，答案上的得数是-1，您确定是对的吗？

追答

对于你输入的题目，我的是正解