Suppose we are given [1-cos(x)]/sin(x). Through calculations, we know that this will be equal to sin(x)/[1+cos(x)]. Now, if we assume an angle value, say x=0.006, they will yield 5.23598532510^-5 and 5.23598776110^-5, respectively. As we can see, there is a discrepancy value of 2.436*10^-11.
My question is, despite the functions being equal, why do their values differ (even though it's small), when we provided an angle value? Can we say that one value is better than the other?
They aren't quite the same. For most values, they are the same (although because of the different method of calculation, a difference due to rounding errors might appear.
However, take for example the value $x=0$. In this case, $\frac{1-\cos(x)}{\sin(x)}$ is undefined, but $\frac{\sin(x)}{1+\cos(x)}$ is zero. However, if you take the limit as $x$ approaches $0$ of $\frac{1-\cos(x)}{\sin(x)}$, you get $\lim_{x\rightarrow0}\frac{1-\cos(x)}{\sin(x)}=\lim_{x\rightarrow0}\frac{\sin(x)}{\cos(x)}=0$. Basically, $\frac{1-\cos(x)}{\sin(x)}$ and $\frac{\sin(x)}{1+\cos(x)}$ are the same everywhere that neither has a removable singularity; if one has a removable singularity, it has a limit of $0$, which is the value of the other at that angle.