[sin2a/(1+cos2a)]*[cosa/(1+cosa)]*cota/2RT.详细过程化简这个式子
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[sin2a/(1+cos2a)]*[cosa/(1+cosa)]*cota/2RT.详细过程化简这个式子
[sin2a/(1+cos2a)]*[cosa/(1+cosa)]*cota/2
RT.详细过程
化简这个式子
[sin2a/(1+cos2a)]*[cosa/(1+cosa)]*cota/2RT.详细过程化简这个式子
原式=[2sinacosa/(1+2cos²a-1)][cosa/(1+cosa)]*cota/2
=(2sinacosa/2cos²a)[cosa/(1+cosa)]*cota/2
=(sina/cosa)[cosa/(1+cosa)]*cota/2
=[sina/(1+cosa)]*cot(a/2)
=[2sin(a/2)cos(a/2)/[1+2cos²(a/2)-1][cos(a/2)/sin(a/2)]
=[sin(a/2)/cos(a/2)][cos(a/2)/sin(a/2)]
=1
(1-sin2a+cos2a)/(1+sin2a-cos2a
化简:(1+sin2a+cos2a)/(1+sin2a-cos2a)
(1+sin2a+cos2a)/(1+sin2a-cos2a)+(1+sin2a-cos2a)/(1+sin2a+cos2a)=
(1-cos2a)/sin2a化简
化简 1-cos2a/sin2a
化简2sin2a/1+cos2a*cos2/cos2a
化简 [(tan2a-1/tan2a)/sin2a-cos2a]+1/cos2a-1/sin2a
求证:1+sin2a-cos2a/1+sin2a+cos2a=tana
证明(1+sin2a-cos2a)/(1+sin2a+cos2a)=tana
化简(sin2a+cos2a-1)(sin2a-cos2a+1)除以sin4a
求证(1+sin2a-cos2a)/(1+sin2a+cos2a)=tan a
1+sin2a-cos2a/1+sin2a+cos2a为什么等于tana?
求证1+sin2a-cos2a除以1+sin2a+cos2a=tana
化简(sin2a+cos2a-1)(sin2a-cos2a+1)除以sin4a
化简:(1+sin2a-cos2a)/(1+sin2a+cos2a)
化简(1+cos2a+sin2a)/(1-cos2a+sin2a)
化简:(1+cos2a)/sin2a-sin2a/(1-cos2a)=
sin2a×cos2a等于1吗