Note that sec6x can be written as [sec2x]3 = [(1 + tan2x)]3
And using the binomial theorem, we have.........
[(1 + tan2x)]3 = [tan2x]3 + 3[tan2x]2 + 3[tan2x] + 1 ... therefore ....
sec6x - tan6x =
[tan2x]3 + 3[tan2x]2 + 3[tan2x] + 1 - tan6x =
tan6x + 3[tan2x]2 + 3[tan2x] + 1 - tan6x =
3[tan4x] + 3[tan2x] + 1 = which is the same thing as the RHS.....