Prove the Trigonometric Identity: $\frac{\sin x}{1-\cos x} = \frac{1+\cos x}{\sin x}$ I'm doing some math exercises but I got stuck on this problem.
In the book of Bogoslavov Vene it says to prove that:
$$\frac{\sin x}{1-\cos x} = \frac{1+\cos x}{\sin x}.$$
It is easy if we do it like this: $\sin^2 x = (1-\cos x)(1+\cos x)=1-\cos^2 x$.
But how to prove it just by going from the left part or from the right
Can anybody help me?
Thank you!
 A: Note that we must have that $\cos x \neq 1$, so x cannot be any even multiple of $\pi$. Similarly, $\sin x \neq 0$, so x cannot be any multiple of $\pi$.
Why do I note the above? 
Because we have an identity, $$\frac{\sin x}{1-\cos x} = \frac{1+\cos x}{\sin x}$$ which is senseless if $\cos x = 1,$ and/or $\sin x = 0$.
$\require{cancel}$
Since $x$ cannot be any multiple of $\pi$, we can multiply numerator and denominator by $1+ \cos x$, because we already ruled out $x$ = integer multiple of $\pi$.
We also use the Pythagorean Identity $$\sin^2x + \cos^2 x =1 \iff 1 - \cos^2 x = \sin^2 x$$

$$\begin{align}
\frac{\sin(x)}{(1-\cos(x))}&= \frac{\sin(x)}{(1-\cos(x))}\cdot \frac{(1+\cos(x))}{(1+\cos(x))} \\ \\ & = \frac{\sin x(1 + \cos x)}{\underbrace{1- \cos^2 x}_{(a + b)(a - b) = a^2 - b^2}} \\ \\
& = \frac {\sin x(1 + \cos x)}{\sin^2 x} \\ \\
& = \frac{\cancel{\sin x}(1 + \cos x)}{\cancel{\sin x}\cdot \sin x} \\ \\
& = \frac{1 +\cos x}{\sin x}\end{align}$$
for all $x \neq k\pi, k\in \mathbb Z$.
A: Note that:
\begin{align*}
\frac{\sin(x)}{(1-\cos(x))}&= \frac{\sin(x)}{(1-\cos(x))} \times \frac{1+\cos(x)}{1+\cos(x)}
\end{align*}
A: $$\sin^2x+\cos^2x=1\iff  \sin^2x=1-\cos^2x=(1+\cos x)(1-\cos x)$$
$$\implies \frac{\sin x}{1-\cos x}=\frac{1+\cos x}{\sin x}$$
