How do you prove: ##sin^2x/(1-cosx) = 1+cosx##?

See the proof in the explanation.

##sin^2x/(1-cosx)##

Multiply the numerator and denominator both by the conjugate of the denominator. This is a "fancy 1" that does not affect the value of the expression.

##sin^2x/(1-cosx)*(1+cosx)/(1+cosx)##

This sets up a Difference of Squares in the denominator. Resist the urge to distribute in the numerator.

##[(sin^2x)(1+cosx)]/((1)^2-(cosx)^2)=[(sin^2x)(1+cosx)]/(1-cos^2x)##

Use the Pythagorean Identity ##sin^2x+cos^2=1## to replace the 1 in your denominator.

##[(sin^2x)(1+cosx)]/((sin^2x+cos^x)-cos^2x)##

Simplify the denominator.

##[(sin^2x)(1+cosx)]/((sin^2x)##

Simplify the fraction.

##1+cosx##

QED.