Question

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

Answer 1

see explanation

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

Answer 2

Alternate solution.

Explanation:

Left hand side (LHS) is

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

We know the identity: `sin^2x+cos^2x=1`

`orsin^2x=1-cos^2x`

Substituting in the numerator of second term
`1 -((1-cos^2x)/(1-cosx))`

Using the factors `x^2-y^2=(x+y)(x-y)`

We obtain
`1 -((cancel((1-cosx))(1+cosx))/cancel(1-cosx))`

Simplifying we obtain
`1-(1+cosx)`
`or cancel1-cancel1-cosx`

`= - cosx =`RHS##

Hence proved