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