Question
Wed August 29, 2012 By:

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

Expert Reply
Wed August 29, 2012
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
Related Questions
Home Work Help