(3-4cos(2x)+cos(4x))/(3+4cos(2x)+cos(4x))=tan^6(x)

Answer 1

For readability, let c=cos(x), s=sin(x)
4cos(2x)+cos(4x) = 4c^2 - 4s^2 + s^4 + c^4 - 6s^2c^2
= (4 - 4sin^2 - 4s^2) + (s^4 + 1 - 2s^2 + s^4 - 6s^2 + 6s^4)
= (4-8s^2) + (1 - 8s^2 + 8s^4)

(3-4cos(2x)+cos(4x)) = 3 - 4 + 8s^2 + 1 - 8s^2 + 8s^4 = 8s^4

(3+4cos(2x)+cos(4x)) = 3 + 4-8s^2 + 1 - 8s^2 + 8s^4 = 8 - 16s^2 + 8s^4 = 8c^4

so, dividing, you get 8s^4/8c^4 = tan^4

You sure you copied it right? Don't see how to make it tan^6

Answer 2

I tested the original equation using x = 10°
the left side was NOT equal to the right side, so the equation is not an identity.

Answer 3

I confirmed the identity, the left hand side IS equal to tan^4(x), not tan^6(x).
A typo is the most likely cause.