求证:sina-cosa+1/sina+cosa-1=tan(a/2+π/4)
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求证:sina-cosa+1/sina+cosa-1=tan(a/2+π/4)
(sina-cosa+1)/(sina+cosa-1)=tan(a/2+π/4)
(sina-cosa+1)/(sina+cosa-1)=tan(a/2+π/4)
因为
(sina-cosa+1)/(sina+cosa-1)-tan(a/2+π/4)
=(sina-cosa+1)/(sina+cosa-1)-[1-cos(a+π/2)]/sin(a+π/2)]
=(sina-cosa+1)/(sina+cosa-1)-(1+sina) / cosa
=[cosa(sina-cosa+1)-(1+sina)(sina+cosa-1)] / [ cosa(sina+cosa-1)]
=[cosasina-cos²a+cosa-(sina+cosa-1+sin²a+cosasina-sina)] / [ cosa(sina+cosa-1)]
=[-cos²a+1-sin²a] / [ cosa(sina+cosa-1)]
=0
左边-右边=0
所以,
(sina-cosa+1)/(sina+cosa-1)=tan(a/2+π/4)成立
(sina-cosa+1)/(sina+cosa-1)-tan(a/2+π/4)
=(sina-cosa+1)/(sina+cosa-1)-[1-cos(a+π/2)]/sin(a+π/2)]
=(sina-cosa+1)/(sina+cosa-1)-(1+sina) / cosa
=[cosa(sina-cosa+1)-(1+sina)(sina+cosa-1)] / [ cosa(sina+cosa-1)]
=[cosasina-cos²a+cosa-(sina+cosa-1+sin²a+cosasina-sina)] / [ cosa(sina+cosa-1)]
=[-cos²a+1-sin²a] / [ cosa(sina+cosa-1)]
=0
左边-右边=0
所以,
(sina-cosa+1)/(sina+cosa-1)=tan(a/2+π/4)成立
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