求证cos2a/(1/tana/2-tana/2)=1/4sin2a

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求证cos2a/(1/tana/2-tana/2)=1/4sin2a
求证cos2a/(1/tana/2-tana/2)=1/4sin2a

求证cos2a/(1/tana/2-tana/2)=1/4sin2a
cos(2a)/[1/tan(a/2)-tan(a/2)]
=cos(2a)/[cot(a/2)-tan(a/2)]
=cos(2a)/{[cos^2(a/2)-sin^2(a/2)]/[sin(a/2)cos(a/2)]}
=cos(2a)/[cosa/(sina/2)]
=sinacos(2a)/(2cosa)
cos2a应该是cos^2(a)
cos^2(a)/[1/tan(a/2)-tan(a/2)]
=sinacos^2(a)/(2cosa)=sinacosa/2=sin(2a)/4

求证：2(sin2a+1)/(1+sin2a+cos2a)=tana+1 求证:1+sin2a-cos2a/1+sin2a+cos2a=tana 求证1+sin2a-cos2a除以1+sin2a+cos2a=tana 求证cos2a/(1/tana/2-tana/2)=1/4sin2a 已知1-tana/2+tana=1,求证3sin2a =-4cos2a 已知1-tanA/2+tanA=1,求证3sin2A=-4cos2A 会的解下已知(1-tanA)/(2+tanA)=1,求证3tan2A=-4cos2A急求证:1+sin2a/cos2a=1+tana/1-tana 求证(1+2sin2a/sin2a+1+cos2a)-(1/2)=(1/2)tana 求证（sin2a+1）/（1+cos2a+sin2a）=1/2tana+1/2 求证:1+2sinacosa/sin2a-cos2a=tana+1/tan-1 1-sin2a/cos2a=1+tan2a/1+tana求证化简(1+cos2a)/(tana/2)-(cota/2) 化简(1+cos2a)/(tana/2-cota/2) 化简（1+cos2a)/(tana/2-1/tana/2) 化简（1+cos2a)/(tana/2-1/tana/2) 求证1＋sin2a－cos2a/1＋sin2a＋cos2a＝tana 求证明：1-sin2a/cos2a = 1-tana/1+tana 和已知tana=2,求sin2a-cos2a/cosa平方.