# Question #a2cb0

Feb 5, 2016

I have to guess here but I'm assuming that you would like to prove the following identity:

$\sin \frac{x}{1 - \cos x} - \frac{1 + \cos x}{\sin} x = 0$

Let's start by bringing the second fraction to the right-hand side:

$\iff \sin \frac{x}{1 - \cos x} = \frac{1 + \cos x}{\sin} x$

Now, in order to get rid of the fractions, you should multiply both sides with both denominators:

$\iff \sin x \cdot \sin x = \left(1 + \cos x\right) \cdot \left(1 - \cos x\right)$

... make use of the formula $\left(a - b\right) \left(a + b\right) = {a}^{2} - {b}^{2}$...

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

... add ${\cos}^{2} x$ on both sides...

$\iff {\sin}^{2} x + {\cos}^{2} x = 1$

However, this is a well-known identity.

As your original equation is equivalent to this identity and the identity certainly holds, your equation holds as well.

q.e.d.