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