化简cos(α+π)tan(2π+α)

2sinα*cosα*cos(2π-α)+cos(π+2α)*cosα*tanα化简! sin3(-α)cos(2π+α)tan(-α-π)化简 化简[sin²(α+π)cos(α+π)]/tan(α+π)cos³(-α-π)tan(-α-2π)化简sin²(α+π)cos(α+π)/tan(α+π)cos³(-α-π)tan(-α-2π) 化简cos(α+π)tan(2π+α) cosαtan（2π+α）化简 化简cos(α-π)sin²(α+2π)/tan (α+4π)tan(α+π)cos³(-α-π) 化简:(sin(α-3π)*cos(-α))/(tan(α-5π)*cos^2(α-5π)*tan(4π-α)) 化简cos(α -π )tan(α-2π)tan(2π-α)/sin(π+α) cos(π-α)tan(2π-α)tan(π-α)/sin(π+α) 化简 化简：[sin(π+α)*cos(π-α)*tan(π-α)]/[cos(π/2+α)*tan(3π/2-α)*tan(α-3π)] 化简sin(π/2+α)cos(3π-α)tan(π+α)/cos(π/2-α)cos(-α-π) 化简{sin(π+α)^3cos(-α)cos(π+α)}/{tan(π+α)^3cos(-π-α)^2} 化简：sin²（α+π)cos(-α+π)/tan(α+π)tan(α+2π)cos²(-α-π)1.cos（27°+α）cos（33°-α）-sin（27°+α）sin（33°-α） 2.sin²（α+π)cos(-α+π)/tan(α+π)tan(α+2π)cos²(-α-π) 不等式证明1已知α、β、γ∈(0,π/2),且tanα+tanβ+tanγ＝3,求证：1/(cosαcosβ)+1/（cosβcosγ)+1/(cosγcosα)≥6. sin3(-α)cos(2π+α)tan(- 2- π） 已知tanα=2,则cos(α+π)等于 sin(2π-α)tan(α+3π)tan(-α-π)/tan(α-3π)cos(π-α) sin²(-α)·cos(π+α)/tan(2∏+α)tan(∏+α)cos³(-∏-α)