证明恒等式 三角比1. sin^2a+sin^2b-sin^2asin^2b+cos^2acos^2b=12. 2(1-sina)(1+cosa)=(1-sina+cosa)^23. (tan^2a-cot^2a)/(sin^2a-cos^2a)=sec^2a+csc^2a过程答案

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

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
=1
2.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

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
=1
2.
左边-右边
=2(1-sina+cosa-sinacosa)-[1+(sina)^2+(cosa)^2 -2sina+2cosa-2sinacosa]
=2(1-sina+cosa-sinacosa)-(2-2sina+2cosa-2sinacosa)
=0
3.
左边=[(sina/cosa)^2-(cosa/sina)^2]/[(sina)^2-(cosa)^2]
=[(sina)^2+(cosa)^2]/[(cosa)^2(sina)^2]
=(seca)^2+(csca)^2