on the right side you can use the pythagorean identity 1 = sin^2 x + cos^2 x and substitute to both the top and the bottom 1's
= [sin^2x + 2 sinx cosx + cos^2x] / [2sin^2x-sin^2x-cos^2x]
by simplifying the top to a perfect square binomial youll get (also combine like terms)
= [sinx +cosx]^2 / [sin^2x-cos^2x]
seperate the bottom binomial
= [sinx +cosx]^2 / [(sinx + cosx)(sinx - cosx)]
cancel one (sinx+cosx) from the top and bottom and your answer is
= [sinx +cosx] / [sinx - cosx] which is equal to the left side
