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

Jun 12, 2016

Modifying just the left-hand side:

We can use the Pythagorean Identity to rewrite ${\sin}^{2} x$. The Pythagorean Identity states that

$\textcolor{red}{\overline{\underline{|}} \textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{{\sin}^{2} x + {\cos}^{2} x = 1} \textcolor{w h i t e}{\frac{a}{a}} |}$

Which can be rearranged to say that

$\textcolor{t e a l}{\overline{\underline{|}} \textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{{\sin}^{2} x = 1 - {\cos}^{2} x} \textcolor{w h i t e}{\frac{a}{a}} |}$

Thus, we see that

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

We can now factor the numerator of this fraction. $1 - {\cos}^{2} x$ is a difference of squares, which can be factored as:

$\textcolor{b l u e}{\overline{\underline{|}} \textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{{a}^{2} - {b}^{2} = \left(a + b\right) \left(a - b\right)} \textcolor{w h i t e}{\frac{a}{a}} |}$

This can be applied to $1 - {\cos}^{2} x$ as follows:

$\textcolor{\mathmr{and} a n \ge}{\overline{\underline{|}} \textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{1 - {\cos}^{2} x = {1}^{2} - {\left(\cos x\right)}^{2} = \left(1 + \cos x\right) \left(1 - \cos x\right)} \textcolor{w h i t e}{\frac{a}{a}} |}$

Therefore,

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

Since there is $1 - \cos x$ present in both the numerator and denominator, it can be cancelled:

$\frac{\left(1 + \cos x\right) \left(1 - \cos x\right)}{1 - \cos x} = \frac{\left(1 + \cos x\right) \textcolor{red}{\cancel{\textcolor{b l a c k}{\left(1 - \cos x\right)}}}}{\textcolor{red}{\cancel{\textcolor{b l a c k}{\left(1 - \cos x\right)}}}} = 1 + \cos x$

This is what we initially set out to prove.