证明恒等式 三角比1. sin^2a+sin^2b-sin^2asin^2b+cos^2acos^2

证明恒等式 三角比1. sin^2a+sin^2b-sin^2asin^2b+cos^2acos^2

1.(sina)^2+(sinb)^2-(sinasinb)^2+(cosacosb)^2=(sina)^2-(sinasinb)^2 + 1-(cosb)^2+(cosacosb)^2=(sina)^2[1-(sinb)^2] - (cosb)^2[1-(cosa)^2] +1=(sinacosb)^2-(cosbsina)^2+1=0+1=12.左边-右边=2(1-sina+cosa-...======以下答案可供参考======供参考答案1:1.左边=sin^2a(1-sin^b)+sin^2b+cos^2acos^2b =sin^2acos^2b+sin^2b+cos^2acos^2b =cos^2b(sin^2a+cos^2a)+sin^2b =12.左边=2(1-sina)+2cosa(1-sina) =2-2sina+2cosa(1-sina) =sin^2a+cos^2a+1-2sina+2cosa(1-sina) =(1-sina)^2+2cosa(1-sina)+cos^2a =(1-sina+cosa)^2 3.左边=（sin^2a/cos^2a-cos^2a/sin^2a）/(sin^2a-cos^2a) =[(sin^4a-cos^4a)/(cos^2asin^2a)]/(sin^2a-cos^2a) =(sin^2a+cos^2a)/(cos^2asin^2a) =1/cos^2a+1/sin^2a=右边供参考答案2:1.sin^2b+sin^2b=1代入右端，消去sin^2b 移项提取公因子得sin^2a*(1-sin^2b)=(1-con^2a)*cos ^2b 等式成立2.左边=2（1-sina+cosa-sina*cosa）右边=1+sin^2a+cos^2a-2sina+2cosa-2sina*cosa=左边3.(sin^2a-cos^2a)*(sec^2a+csc^2a) =tan^2a-cot^2a+1-1=tan^2a-cot^2a

就是这个解释