Solve identity in terms of cos

You should aware that cos2x = 2cos^2 x - 1 and cos3x = 4cos^3 x - cosx
Hence cos^2 x = (1 + cos2x) / 2 and cos^3 x = (3cosx + cos3x) / 4

So
[(sin^4)x][(tan^2)x]
= sin^4 x ( sin^2 x / cos^2 x )
= sin^6 x / cos^2 x
= (1 - cos^2 x)^3 / cos^2 x
= (1 - cos2x)^3 / 4(1 + cos2x)
= (1 - 3cos2x + 3cos^2 2x - cos^3 2x) / 4(1 + cos2x)
= [1 - 3cos2x + 3(1 + cos4x) / 2 - (3cos2x + cos6x) / 4] / 4(1 + cos2x)
= (10 - 15cos2x + 6cos4x - cos6x) / 16(1 + cos2x)

Expand...
(1 - cos^2 x)^3 = 1 - 3 cos^2 x + 3 cos^4 x - cos^6 x