cos(α-π/2)*tan(-α-3π/2)*sec(-α+5π/2)*tan(α+5π/2)

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. 化简:(sin(α-3π)*cos(-α))/(tan(α-5π)*cos^2(α-5π)*tan(4π-α)) sin(2π-α)tan(α+3π)tan(-α-π)/tan(α-3π)cos(π-α) 化简：[sin(π+α)*cos(π-α)*tan(π-α)]/[cos(π/2+α)*tan(3π/2-α)*tan(α-3π)] 求证：[tan（2π-α）cos（3π/2-α)cos(6π-α)]/[sin(α+3π/2)cos(α+3π/2)]=-tanα 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} 已知tan(π+α)=-1/3 tan(α+β)=[sin(π-2α)+4(cosα)^2]/[10(cosα)^2-sin2α]求……已知tan(π+α)=-1/3tan(α+β)=[sin(π-2α)+4(cosα)^2]/[10(cosα)^2-sin2α]求：1、tan(α+β)2、tanβ 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) sin3(-α)cos(2π+α)tan(- 2- π） 化简cos(α+π)tan(2π+α) cosαtan（2π+α）化简 已知tanα=2,则cos(α+π)等于 化简cos(α-π)sin²(α+2π)/tan (α+4π)tan(α+π)cos³(-α-π) cos(α-π/2)*tan(-α-3π/2)*sec(-α+5π/2)*tan(α+5π/2) 计算sin(-α-5π)*cos(α-2分之π)-tan(α-2分之3π)*tan(2π-α)