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

see explanation
To prove , require to manipulate one of the sides into the form of the other.
choosing the left side (LHS) gives
## 1 -(sin^2x/(1-cosx))##
require to combine these : rewrite ## 1 = (1-cosx)/(1-cosx) ##
now have : ## (1-cosx)/(1-cosx) – sin^2x/(1-cosx) ##
basically subtracting 2 fractions with a common denominator
hence ##( 1-cosx-sin^2x)/(1-cosx) = ((1-sin^2x) -cosx)/(1-cosx)##
now ## sin^2x + cos^2x = 1 rArr cos^2 x = 1 -sin^2x##
replacing this result into numerator
to obtain : ##( cos^2x – cosx)/(1 – cosx) ##
‘taking’ out a common factor of -cosx
## =( -cosx( 1 – cosx))/(1-cosx)##
## rArr -cosxcancel(1-cosx)/cancel(1-cosx) = – cosx ##= RHS
thus proved