求证(1+sinθ+cosθ)/(1+sinθ-cosθ)+(1-cosθ+sinθ)/(1+cosθ+sinθ)=2/sinθ如题.请写出求证过程,用高一(下)三角恒等变换知识来证.
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求证(1+sinθ+cosθ)/(1+sinθ-cosθ)+(1-cosθ+sinθ)/(1+cosθ+sinθ)=2/sinθ
如题.请写出求证过程,用高一(下)三角恒等变换知识来证.
(1+sinθ+cosθ)/(1+sinθ-cosθ)=[2sin(θ/2)cos(θ/2)+2cos²(θ/2)]/[2sin(θ/2)cos(θ/2)+2sin²(θ/2)]=cos(θ/2)/sin(θ/2).
(1-cosθ+sinθ)/(1+cosθ+sinθ)=[2sin(θ/2)cos(θ/2)+sin²(θ/2)]/[2sin(θ/2)cos(θ/2)+2cos²(θ/2)]=sin(θ/2)/cos(θ/2).
所以,原式=cos(θ/2)/sin(θ/2)+sin(θ/2)/cos(θ/2)=[cos²(θ/2)+sin²(θ/2)]/[sin(θ/2)cos(θ/2)]=2/sinθ.
原式=(1+sinθ+cosθ)^2/[(1+sinθ)^2-(cosθ)^2]+(1-cosθ+sinθ)^2/[(1+sinθ)^2-(cosθ)^2]=2*(1+sinθ)^2+2(cosθ^2)/[(1+sinθ)^2-(cosθ)^2]= [2+4sinθ+2(sinθ^2)+2(cosθ)^2 ]/ 2sinθ^2 +2sinθ = 2/sinθ
这种证明题关键是在分子分母同时乘以分子对分母进行有理化,剩下的就是化简。
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