Answer
We want to verify that $\frac{1 - cos\ a}{sin\ a}=\frac{sin\ a}{1+cos\ a}$
Start with the right-hand side:
$\frac{1 - cos\ a}{sin\ a}$, we will multiply by $1=\frac{sin\ a}{1+cos\ a}\frac{1+cos\ a}{sin\ a}$
$=\frac{1 - cos\ a}{sin\ a}\frac{sin\ a}{1+cos\ a}\frac{1+cos\ a}{sin\ a}$, where $(1 - cos\ a)(1 + cos\ a)=(1 - cos^2\ a)$ and $(sin\ a)(sin\ a)=sin^2\ a$
$=\frac{1 - cos^2\ a}{sin^2\ a}\frac{sin\ a}{1+cos\ a}$, where we know that $1 - cos^2\ a=sin^2\ a$
$=\frac{sin^2\ a}{sin^2\ a}\frac{sin\ a}{1+cos\ a}\frac{1}{}$, where we can cancel $\frac{sin^2\ a}{sin^2\ a}$
$=\frac{sin\ a}{1+cos\ a}$, which is equal to the left-hand side.
Work Step by Step
We want to verify that $\frac{1 - cos\ a}{sin\ a}=\frac{sin\ a}{1+cos\ a}$
Start with the right-hand side:
$\frac{1 - cos\ a}{sin\ a}$, we will multiply by $1=\frac{sin\ a}{1+cos\ a}\frac{1+cos\ a}{sin\ a}$
$=\frac{1 - cos\ a}{sin\ a}\frac{sin\ a}{1+cos\ a}\frac{1+cos\ a}{sin\ a}$, where $(1 - cos\ a)(1 + cos\ a)=(1 - cos^2\ a)$ and $(sin\ a)(sin\ a)=sin^2\ a$
$=\frac{1 - cos^2\ a}{sin^2\ a}\frac{sin\ a}{1+cos\ a}$, where we know that $1 - cos^2\ a=sin^2\ a$
$=\frac{sin^2\ a}{sin^2\ a}\frac{sin\ a}{1+cos\ a}\frac{1}{}$, where we can cancel $\frac{sin^2\ a}{sin^2\ a}$
$=\frac{sin\ a}{1+cos\ a}$, which is equal to the left-hand side.
