(3-4sin^2thita)(1-3tan^A)=(3-tan^2A)(4cos^2A-3)

Answer 1

Help please to solved this question?

Answer 2

I will assume there is a typo and angles are A
(3-4sin^2A)(1-3tan^A)=(3-tan^2A)(4cos^2A-3)
I find it that in most cases switching everything to sines and cosines works

(3 - 4sin^2 A)(1 - 3sin^2 A/cos^2 a) = (3 - sin^2 A/cos^2 A)(4cos^2 A - 3)
Now only cosines:
(3 - 4(1 - cos^2 A))(1 - 3(1 - cos^2 A)/cos^2 A) = (3 - (1-cos^2 A)/cos^2 A)(4cos^2 A - 3)
let x = cos^2 A
(3 - 4(1 - x) )(1 - 3(1-x)/x ) = (3 - (1 - x)/x )(4x -3)
(4x - 3)(4x - 3)/x = (4x - 1)/x ( 4x - 3)
multiply both sides by x/(4x-3)
4x-3 = 4x-1

no solution, but please check my algebra steps

Answer 3

It turns out to be an identity.
Both sides expand out to be
-4sin^2A - 9tan^2A + 12sin^2Atan^2A + 3

Answer 4

I became suspicious that it might be an identity when my last line was
(4x - 3)(4x - 3)/x = (4x - 1)/x ( 4x - 3)
my algebra error carried right down to that line