How do you prove the identity (sinx – cosx)/(sinx + cosx) = (2sin^2x-1)/(1+2sinxcosx)?

Use the Pythagorean identity to manipulate the right-hand side: since ##sin^2x+cos^2x=1## we know that ##-1=-sin^2x-cos^2x##. We can plug these in for the ##-1## and ##1## in the equation.
##= (2 sin^2 x-sin^2 x – cos^2 x)/(sin^2 x + cos^2 x+ 2 sin x cos x)##
In the numerator combine the like terms. In the denominator recognize that this is in the pattern for a perfect square binomial: ##(a+b)^2=a^2+2ab+b^2##. Here ##a=sinx## and ##b=cosx##.
##=(sin^2 x – cos^2 x)/(sin x + cos x )^2##
From here the numerator can be factored as a difference of squares which takes the form: ##a^2-b^2=(a+b)(a-b)##. Again ##a=sinx## and ##b=cosx##.
##= ((sin x + cos x)(sin x – cos x))/(sin x + cos x)^2##
Cancel the ##(sinx+cosx)## term.
##= (sin x – cos x)/(sinx + cos x)##
This is the left-hand side of the equation.