求证 1/sin2x+1/cos2x-1/tan2x=2+tan2x
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求证 1/sin2x+1/cos2x-1/tan2x=2+tan2x
2为次数
2为次数
![求证 1/sin2x+1/cos2x-1/tan2x=2+tan2x](/uploads/image/z/19762470-54-0.jpg?t=%E6%B1%82%E8%AF%81+1%2Fsin2x%2B1%2Fcos2x-1%2Ftan2x%3D2%2Btan2x)
证明:
1/sin^2 x+1/cos^2 x-1/tan^2 x
=(sin^2 x+cos^2 x)/sin^2 x+(sin^2 x+cos^2 x)/cos^2 x - 1/tan^2 x
=1+1/tan^2 x+1+tan^2 x-1/tan^2 x
=2+tan^2 x
1/sin^2 x+1/cos^2 x-1/tan^2 x
=(sin^2 x+cos^2 x)/sin^2 x+(sin^2 x+cos^2 x)/cos^2 x - 1/tan^2 x
=1+1/tan^2 x+1+tan^2 x-1/tan^2 x
=2+tan^2 x
(cos2x-sin2x)/[(1-cos2x)(1-tan2x)] =cos2x/(1-cos2x)=[cosx)^2
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