已知cos^4β/cos^2a+sin^4β/sin^2a=1,求证cos^4a/cos^2β+sin^4a/sin2^β=1跪谢,我一定会追加50分的!

已知cos^4β/cos^2a+sin^4β/sin^2a=1,求证cos^4a/cos^2β+sin^4a/sin2^β=1跪谢,我一定会追加50分的!
已知cos^4β/cos^2a+sin^4β/sin^2a=1,求证cos^4a/cos^2β+sin^4a/sin2^β=1
跪谢,我一定会追加50分的!

已知cos^4β/cos^2a+sin^4β/sin^2a=1,求证cos^4a/cos^2β+sin^4a/sin2^β=1跪谢,我一定会追加50分的!
证明：
cos^4β/cos^2a+sin^4β/sin^2a=1
cos^4βsin²α+sin^4βcos²α=cos²αsin²α
(1-sin²β)cos²βsin²α+(1-cos²β)sin²βcos²α=cos²αsin²α
cos²βsin²α+sin²βcos²α-sin²βcos²βsin²α-cos²βsin²βcos²α=cos²αsin²α
cos²βsin²α+sin²βcos²α-sin²βcos²β(sin²α+cos²α)=cos²αsin²α
cos²βsin²α+sin²βcos²α-sin²βcos²β=cos²αsin²α
cos²βsin²α+sin²βcos²α-sin²βcos²β-cos²αsin²α=0
cos²β(sin²α-sin²β)-cos²α(sin²α-sin²β)=0
(cos²β-cos²α)(sin²α-sin²β)=0
所以cos²β=cos²α,sin²α=sin²β
所以
cos^4a/cos^2β+sin^4a/sin2^β
=(cos²β)²/cos²β+(sin²β)²/sin²β
=cos²β+sin²β
=1

cos^4β/cos^2a+sin^4β/sin^2a=1,
所以 两边同时乘以cos^2a×sin^2a
得sin^2a cos^4β+sin^4βcos^2a=cos^2a×sin^2a
求证cos^4a/cos^2β+sin^4a/sin2^β=1
可证 两边同时乘以sin2^β×cos^2β
cos^4a sin2^β+sin^4a cos^...

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cos^4β/cos^2a+sin^4β/sin^2a=1,
所以 两边同时乘以cos^2a×sin^2a
得sin^2a cos^4β+sin^4βcos^2a=cos^2a×sin^2a
求证cos^4a/cos^2β+sin^4a/sin2^β=1
可证 两边同时乘以sin2^β×cos^2β
cos^4a sin2^β+sin^4a cos^2β =sin2^β×cos^2β
因为sin^2a cos^4β+sin^4βcos^2a=cos^2a×sin^2a
恒成立
所以与角度大小无关
所以 a 、β互换原式仍成立
所以cos^4a sin2^β+sin^4a cos^2β =sin2^β×cos^2β
cos^4a/cos^2β+sin^4a/sin2^β=1成立
累死我了

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若cos^2β＝cos^2a，则sin^2β＝sin^2a，结论显然成立
若cos^2β≠cos^2a，cos^2a-cos^2β＝sin^2β-sin^2a
cos^4β/cos^2a+sin^4β/sin^2a=1=cos^2β+sin^2β
cos^2β/cos^2a(cos^2β-cos^2a)=sin^2β/sin^2a(sin^2a-sin^2β)
co...

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若cos^2β＝cos^2a，则sin^2β＝sin^2a，结论显然成立
若cos^2β≠cos^2a，cos^2a-cos^2β＝sin^2β-sin^2a
cos^4β/cos^2a+sin^4β/sin^2a=1=cos^2β+sin^2β
cos^2β/cos^2a(cos^2β-cos^2a)=sin^2β/sin^2a(sin^2a-sin^2β)
cos^2β/cos^2a(cos^2β-cos^2a)=sin^2β/sin^2a(cos^2β-cos^2a)
cos^2β/cos^2a=sin^2β/sin^2a
cos^2a/cos^2β(cos^2a-cos^2β)=sin^2a/sin^2β(cos^2a-cos^2β)
cos^2a/cos^2β(cos^2a-cos^2β)=sin^2a/sin^2β(sin^2β-sin^2a)
cos^4a/cos^2β-cos^2a=sin^2a-sin^4a/sin^2β
cos^4a/cos^2β+sin^4a/sin^2β=sin^2a+cos^2a=1

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