求证:1+sina+cosa/1+sina-cosa+1-cosa+sina/1+cosa+sina=2/sina

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求证:1+sina+cosa/1+sina-cosa+1-cosa+sina/1+cosa+sina=2/sina
求证:1+sina+cosa/1+sina-cosa+1-cosa+sina/1+cosa+sina=2/sina

求证:1+sina+cosa/1+sina-cosa+1-cosa+sina/1+cosa+sina=2/sina
1+sina+cosa/1+sina-cosa+1-cosa+sina/1+cosa+sina
=[(1+sina+cosa)²+(1+sina-cosa)²]/[(1+sina)²-cos²a]
=(1+sin²a+cos²a+2sina+2cosa+2sinacosa+1+sin²a+cos²a+2sina-2cosa-2sinacosa)/[(1+sina)²-cos²a]
=(4+4sina)/[(1+sina)²-(1-sin²a)]
=4(1+sina)/(1+sina)(1+sina-1+sina)
=4/2sina
=2/sina