Question

Sec^2a-(sin^2a-2sin^4a)/(2cos^4a-cos^2a) =1

Answer 1

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