cos(α -π )tan(α-2π)tan(2π-α)/sin(π+α)
=[-cos(α)]tan(α)[-tan(α)]/[-sin(α)]
=tan(α)[-tan(α)]/[tan(α)]
= -tan(α)
化简cos(α -π )tan(α-2π)tan(2π-α)/sin(π+α)
1个回答
相关问题
-
化简：sin²（α+π)cos(-α+π)/tan(α+π)tan(α+2π)cos²(-α-π)
-
化简：sin2(α+π)•cos(π+α)tan(π+α)•cos3(−α−π)•tan(−α−2π)．
-
[sin(2π+α）·tan(-α)·cos(-α)]÷[sin(-α)·cos(2π+α）·tan(2π+α）化简
-
sin｛π－α｝cos｛2π－α｝tan｛－α＋π｝/sin｛π＋α｝化简
-
化简cos(π-α)sin(α+2π)/cos^2(-α-π)tan(2π-α
-
化简：（Ⅰ）sin(α−2π)cos(α+π)tan(α−99π)cos(π−α)sin(3π−α)sin(−α−π)；
-
化简（1）tan(π−α)cos(2π−α)sin(−α+3π2)cos(−α−π)sin(−π−α)．
-
化简：（1）tan(π−α)•sin2(α+π2)•cos(2π−α)cos3(−π−α)•tan(α−2π)；（2）s
-
化简（1）sin(α+π)cos(-α)sin(-α-π) （2）sin³（-α）cos（2π+α）tan（-
-
化简：tan(π+α)cos(2π+α)sin(α−3π2)cos(−α−3π)sin(−3π−α)=______．