3-4sin^2a/cos^a=3-tan^2a
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can i know what to prove please
got it but I don't think it is right but it may help you to solve it
LHS
(3-4sin^2a)/cos^a
=(3-3sin^2a-sin^2a)/cos^a
=(3{1-sin^2a}-sin^2a)/cos^a
=(3cos^2a-sin^2a)/cos^a
=(3cos^2a/cos^a)- (sin^2a/cos^a)
now divide whole eq with cos^a
=3-tan^2a
got it but I don't think it is right but it may help you to solve it
LHS
(3-4sin^2a)/cos^a
=(3-3sin^2a-sin^2a)/cos^a
=(3{1-sin^2a}-sin^2a)/cos^a
=(3cos^2a-sin^2a)/cos^a
=(3cos^2a/cos^a)- (sin^2a/cos^a)
now divide whole eq with cos^a
=3-tan^2a
MdNeyazAhmed:
LHS = RHS
tq
ok anyone problem
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