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.