How to prove the identity sin^4x - cos^4x / sin^3x - cos^3x = Sinx + cosx / 1 + sinxcosx

We have to prove that: [(sin x)^4 - (cos x)^4]/[(sin x)^3
- (cos x)^3 = (sin x + cos x) / (1 + sin x* cos x)

[(sin
x)^4 - (cos x)^4]/[(sin x)^3 - (cos x)^3

use a^2 - b^2 = (a
- b)(a + b) and a^3 - b^3 = (a - b)*(b^2  + a*b +
a^2)

=> [(sin x - cos x)(sin x + cos x)(sin x)^2 +
(cos x)^2]/(sin x - cos x)((sin x)^2 + sin x*cos x + (cos
x)^2)

=> [(sin x + cos x)(sin x)^2 + (cos x)^2]/(sin
x - cos x)((sin x)^2 + sin x*cos x + (cos x)^2)

use (sin
x)^2 + (cos x)^2 = 1

=> (sin x + cos x)*1/(sin x -
cos x)(1 + sin x*cos x)

This proves: [(sin
x)^4 - (cos x)^4]/[(sin x)^3 - (cos x)^3 = (sin x + cos x) / (1 + sin x* cos
x)