证明(tan^2x-cot^2x)/(sin^2x+cos^2x)=sec^2x+csc^2x
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证明(tan^2x-cot^2x)/(sin^2x+cos^2x)=sec^2x+csc^2x
sin^2x+cos^2x=1
所以左边=tan^2x-cot^2x
=sin^2x/cos^2x-cos^2x/sin^2x
=(sin^4x-cos^4x)/sin^2xcos^2x
=(sin^2x+cos^2x)(sin^2x-cos^2x)/sin^2xcos^2x
=(sin^2x-cos^2x)/sin^2xcos^2x
=sin^2x/sin^2xcos^2x-cos^2x/sin^2xcos^2x
=1/cos^2x-1/sin^2x
=sec^2x-csc^2x
右边是不是写错了?
所以左边=tan^2x-cot^2x
=sin^2x/cos^2x-cos^2x/sin^2x
=(sin^4x-cos^4x)/sin^2xcos^2x
=(sin^2x+cos^2x)(sin^2x-cos^2x)/sin^2xcos^2x
=(sin^2x-cos^2x)/sin^2xcos^2x
=sin^2x/sin^2xcos^2x-cos^2x/sin^2xcos^2x
=1/cos^2x-1/sin^2x
=sec^2x-csc^2x
右边是不是写错了?
证明(tan^2x-cot^2x)/(sin^2x-cos^2x)=sec^2x+csc^2x
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