solve the given question
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This is the second part of the question
Sin^4/a+cos^4/b=1/a+b
(a+b)[sin^4/a+ cos^4/b]=(1^2)
Sin^4+sin^4b/a+cos^4+cos^4a/b
= (sin^2+cos^2)^2
Hence then
rootb/asin^2-roota/bcos^2=0
Sin^4/a+cos^4/b=1/a+b
(a+b)[sin^4/a+ cos^4/b]=(1^2)
Sin^4+sin^4b/a+cos^4+cos^4a/b
= (sin^2+cos^2)^2
Hence then
rootb/asin^2-roota/bcos^2=0
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Jindalnayan:
Thanks
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