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

Feb 9, 2016

see explanation

#### Explanation:

To prove , require to manipulate one of the sides into the form of the other.
choosing the left side (LHS) gives

$1 - \left({\sin}^{2} \frac{x}{1 - \cos x}\right)$

require to combine these : rewrite $1 = \frac{1 - \cos x}{1 - \cos x}$

now have : $\frac{1 - \cos x}{1 - \cos x} - {\sin}^{2} \frac{x}{1 - \cos x}$

basically subtracting 2 fractions with a common denominator

hence $\frac{1 - \cos x - {\sin}^{2} x}{1 - \cos x} = \frac{\left(1 - {\sin}^{2} x\right) - \cos x}{1 - \cos x}$

now ${\sin}^{2} x + {\cos}^{2} x = 1 \Rightarrow {\cos}^{2} x = 1 - {\sin}^{2} x$

replacing this result into numerator

to obtain : $\frac{{\cos}^{2} x - \cos x}{1 - \cos x}$

'taking' out a common factor of -cosx

$= \frac{- \cos x \left(1 - \cos x\right)}{1 - \cos x}$

$\Rightarrow - \cos x \frac{\cancel{1 - \cos x}}{\cancel{1 - \cos x}} = - \cos x$= RHS

thus proved

Feb 9, 2016

Alternate solution.

#### Explanation:

Left hand side (LHS) is

$1 - \left({\sin}^{2} \frac{x}{1 - \cos x}\right)$

We know the identity: ${\sin}^{2} x + {\cos}^{2} x = 1$

$\mathmr{and} {\sin}^{2} x = 1 - {\cos}^{2} x$

Substituting in the numerator of second term
$1 - \left(\frac{1 - {\cos}^{2} x}{1 - \cos x}\right)$

Using the factors ${x}^{2} - {y}^{2} = \left(x + y\right) \left(x - y\right)$

We obtain
$1 - \left(\frac{\cancel{\left(1 - \cos x\right)} \left(1 + \cos x\right)}{\cancel{1 - \cos x}}\right)$

Simplifying we obtain
$1 - \left(1 + \cos x\right)$
$\mathmr{and} \cancel{1} - \cancel{1} - \cos x$

$= - \cos x =$RHS

Hence proved