sec^2 A-[(sin^2 A-2sin^4 A)/(2cos^4 A-cos^2 A)]=1

Asked by  | 29th Aug, 2012, 09:21: PM

Expert Answer:

LHS = 
sec^2 A-[(sin^2 A-2sin^4 A)/(2cos^4 A-cos^2 A)]
= sec^2 A-[sin^2 A(1-2sin^2 A)/cos^2 A(2cos^2 A-1)]
= sec^2 A-[sin^2 A(cos^A-sin^2 A)/cos^2 A(cos^2 A-sin^2A)]
= sec^2 A-[sin^2 A(cos^A-sin^2 A)/cos^2 A(cos^2 A-sin^2A)]
= sec^2 A-tan^2A
= 1
= RHS

Answered by  | 29th Aug, 2012, 11:27: PM

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