化简(1+cosa+sina)/(1+sina-cosa)
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化简(1+cosa+sina)/(1+sina-cosa)
原式=(1+sinA+cosA)/(1+sinA-cosA)
=(1+cosA+sinA)/[(1-cosA)+sinA]
={2[cos(A/2)]^2+2sin(A/2)cos(A/2)}
/{2[sin(A/2)]^2+2sin(A/2)cos(A/2)|
=2(cos(A/2)[cos(A/2)+sin(A/2)]/{2sin(A/2)[sin(A/2)+cos(A/2)]|
=cos(A/2/sin(A/2)
=cot(A/2).
=(1+cosA+sinA)/[(1-cosA)+sinA]
={2[cos(A/2)]^2+2sin(A/2)cos(A/2)}
/{2[sin(A/2)]^2+2sin(A/2)cos(A/2)|
=2(cos(A/2)[cos(A/2)+sin(A/2)]/{2sin(A/2)[sin(A/2)+cos(A/2)]|
=cos(A/2/sin(A/2)
=cot(A/2).
化简(sina+cosa-1)(sina-cosa+1)/sin2a
化简(1+cosa+sina)/(1+sina-cosa)
化简(1+sina-cosa)/(1+sina+cosa)
化简cosa*根号下(1-sina/1+sina)+sina*根号下(1-cosa/1+cosa)
(1+sina-cosa)/(1-sina+cosa)等于多少?
(cosa-sina)/(cosa+sina)=(1-tana)/(1+tana)
化简,1-sinA/cosA
1-sina/cosa 化简
化简,(1+sina-cosa)/(1+sina+cosa)+(1+sina+cosa)/(1+sina-cosa).答
化简1+cosa-sina/1-cosa-sina+1-cosa-sina/1+cosa-sina(要过程)
同角三角函数 习题证明2(cosa-sina )/1+cosa+sina=cosa/1+sina-sina/1+cosa
求证:1+sina+cosa/1+sina-cosa+1-cosa+sina/1+cosa+sina=2/sina