求证tan(2π-a)cos(3π/2 -a)cos(6π -a) / sin(a+3π/2)cos(a+3π/2) = -tan求证tan(2π-a)cos(3π/2 -a)cos(6π -a) / sin(a+3π/2)cos(a+3π/2) = -tan
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求证tan(2π-a)cos(3π/2 -a)cos(6π -a) / sin(a+3π/2)cos(a+3π/2) = -tan求证tan(2π-a)cos(3π/2 -a)cos(6π -a) / sin(a+3π/2)cos(a+3π/2) = -tan
求证tan(2π-a)cos(3π/2 -a)cos(6π -a) / sin(a+3π/2)cos(a+3π/2) = -tan
求证tan(2π-a)cos(3π/2 -a)cos(6π -a) / sin(a+3π/2)cos(a+3π/2) = -tan
求证tan(2π-a)cos(3π/2 -a)cos(6π -a) / sin(a+3π/2)cos(a+3π/2) = -tan求证tan(2π-a)cos(3π/2 -a)cos(6π -a) / sin(a+3π/2)cos(a+3π/2) = -tan
tan(2π-a)cos(3π/2 -)cos(6π -a) / sin(+3π/2)cos(+3π/2) =(-tan)(-cosa)cosa/(-cosa)cosa=-tana
求证tan(2π-a)cos(3π/2 -a)cos(6π -a) / sin(a+3π/2)cos(a+3π/2) = -tan求证tan(2π-a)cos(3π/2 -a)cos(6π -a) / sin(a+3π/2)cos(a+3π/2) = -tan
求证:2sin(a-3π/2)cos(a+π/2)-1/1-2sin^(π+a)=tan(9π+a)+1/tan(π+a)-1
求证:3-2cos^2a=3tan^2a+1/tan^2a+1
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.
[sin^2(-a)*cos(π+a)]/[tan(2π+a)*tan(π+a)*cos^3(-π-a)]=?
求证:[tan(2π-α)cos(3π/2-α)cos(6π-α)]/[sin(α+3π/2)cos(α+3π/2)]=-tanα
求证:tanα+1/tan(π/4+α/2)=1/cosα帮个忙!
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求证tan(3a/2)-tan(a/2)=【2sin(a/2)】/【cos(3a/2)】
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求证:tan(2π-a) sin(-2π-a) cos(6π-a) / cos(a-π) sin(5π-a)=-tana如题,
求证:tan(2π-a)sin(-2π-a)cos(6π-a)/cos(a-π)sin(5π-a)=-tana
求证(cos^2 A)/{[1/tan(a/2)]-tan(A/2)=(sin2A)/4
求证(1-tan^4A)cos^2A+tan^2A=1..