I think your question means (sin^(2)x)/(1+cosx), if yes then your answer is:
so the trig identity sin^(2) x = 1 - cos^(2)x , which is the form of expression of a^2 - b^2, which by factoring equals (a-b)(a+b).
So 1-cos^(2)x will become (1-cosx)(1+cosx)
Your final fraction would be [(1-cosx)(1+cosx)]/(1+cosx), where 1+cosx will cancel itself on the numerator and the denominator, thus your answer will be 1-cosx