Rewrite the following expression in terms of tan3x: (sin2x + sin4x)/(cos2x + cos4x)

sin2x+sin4x = 2sin3x cosx

cos2x+cos4x = 2cos3x cosx

Now divide to get tan3x

To rewrite the expression (sin2x + sin4x)/(cos2x + cos4x) in terms of tan3x, we can make use of the tangent double-angle formula:

tan(2θ) = (2tanθ)/(1 - tan²θ)

First, let's write sin2x and cos2x in terms of tanx:

sin2x = 2sinx*cosx
cos2x = cos²x - sin²x = cos²x - (1 - cos²x) = 2cos²x - 1
tan2x = (2sinx*cosx)/(2cos²x - 1)

Similarly, let's write sin4x and cos4x in terms of tan2x:

sin4x = 2sin2x*cos2x = 2(2sinx*cosx)(2cos²x - 1)
cos4x = cos²2x - sin²2x = cos²2x - (1 - cos²2x) = 2cos²2x - 1
tan4x = (2(2sinx*cosx)(2cos²x - 1))/(2cos²2x - 1)

Now, substitute these expressions back into the original expression:

(sin2x + sin4x)/(cos2x + cos4x) = (2sinx*cosx + 2(2sinx*cosx)(2cos²x - 1))/(2cos²x - 1 + 2cos²2x - 1)

Now, we can simplify the expression:

= (2sinx*cosx(1 + 4(2cos²x - 1))/(2cos²x + 2cos²2x - 2)

= (2sinx*cosx(1 + 8cos²x - 4))/(2cos²x + (2cos²x - 1)²)

= (2sinx*cosx(8cos²x - 3))/(2cos²x + 4cos⁴x - 4cos²x + 1)

= (16sinx*cos²x - 6sinx*cosx)/(4cos⁴x - 2cos²x + 2cos²x + 1)

= 16sinx*cos²x - 6sinx*cosx)/(4cos⁴x + 1)

Finally, substituting tanx = sinx/cosx, we can express the expression in terms of tan3x:

= 16(tanx)(1 - tan²x)/(4(1 - tan²x)² + 1)

Therefore, the expression (sin2x + sin4x)/(cos2x + cos4x) can be rewritten in terms of tan3x as:

16(tan3x)/(4(1 - tan²3x)² + 1)