Why isn't this square root $+$ or $-$? I was tasked with proving the identity $\tan(\frac x 2) = \dfrac {\sin(x)}{1+\cos(x)}$
I used the quotient identity for tangent and the half angle identities for sine and cosine to get $ \pm \dfrac {\sqrt{\dfrac {1-\cos(x)}{2}}}{\sqrt{\dfrac {1-\cos(x)}{2}}}$
which I reduced to $\pm \sqrt{\dfrac {1-\cos(x)}{1+\cos(x)}}$
I multiplied the fraction (within the square root) by $ \dfrac {1+ \cos(x)}{1+\cos(x)}$
Resulting in $\pm \sqrt{\dfrac {1-\cos^{2}(x)}{(1+\cos(x))^2}}$
Using the Pythagorean identity, I get $\pm\sqrt{\dfrac {\sin^{2}(x)}{(1+\cos(x))^2}}$
Taking the square root of the numerator and denominator I further reduced to $\pm \dfrac {\sin(x)}{1+\cos(x)}$
I thought I was done but when I checked my work in the answer book, it showed $ \left|\dfrac {\sin(x)}{1+\cos(x)}\right|$
Where do they get the absolute value from? 
 A: Avoid unnecessary squaring wherever possible.
If we use the following approach, no such confusion arises.
$$\tan\frac{x}{2}=\frac{\sin\frac{x}{2}}{\cos\frac{x}{2}}$$
Now multiply numerator & denominator by $2\cos\frac{x}{2}$
$$\frac{\sin\frac{x}{2}}{\cos\frac{x}{2}}=\frac{2\cos\frac{x}{2}\sin\frac{x}{2}}{2\cos^2\frac{x}{2}}.$$
Now $\sin2A=2\sin A\cos A$   and $\cos2A=2\cos^2A-1$
So, $\tan\frac{x}{2}=\frac{\sin x}{1+\cos x}$
If the multiplier is $2\sin\frac{x}{2}$,  $\tan\frac{x}{2}$ will be $\frac{1-\cos x}{\sin x}$ which is same as $\frac{\sin x}{1+\cos x}$

If we follow your approach also, just observe that $1+\cos x$ can not be negative as $-1≤\cos A≤1$  and $\frac{\tan\frac{x}{2}}{\sin x}=\frac{\sin\frac{x}{2}}{\cos\frac{x}{2}2\cos\frac{x}{2}\sin\frac{x}{2}}=\frac{1}{2\cos^2\frac{x}{2}}$ which also $>0$.
So the sign of $\frac{\tan\frac{x}{2}}{\frac{\sin x}{1+\cos x}}$ is positive.
So the sign of $\tan\frac{x}{2}$ and $\frac{\sin x}{1+\cos x}$ are same.
A: You are tasked with proving that $\tan(\frac x 2) = \dfrac {\sin(x)}{1+\cos(x)}$.  If we suppose that this is not an impossible task (i.e. that the identity is correct) then neither $\tan(\frac x 2) =\pm\dfrac {\sin(x)}{1+\cos(x)}$ nor $\tan(\frac x 2) =\left|\dfrac {\sin(x)}{1+\cos(x)}\right|$ are satisfactory places to end up.  The first version, which you arrived at, is not exactly incorrect, but rather incomplete, because it doesn't say which sign applies.  The second version, from the answer book, is incorrect when $\tan(\frac x2)<0$.
In general, $a=\pm b$ gives you the same information as $|a|=|b|$.  However, in this case we can determine which sign applies.  Your work essentially shows that $\left|\tan(\frac x 2)\right| =\left|\dfrac {\sin(x)}{1+\cos(x)}\right|$.  But note that $1+\cos(x)\geq 0$ so it does not affect the sign, and $\sin(x)$ always has the same sign as $\tan(x/2)$.  To see that the last point is true, it is enough to work in the interval $(-\pi,\pi)$ by periodicity.  Both $\tan(x/2)$ and $\sin(x)$ are positive when $x$ is in $(0,\pi)$, and both are negative when $x$ is in $(-\pi,0)$.
That said, I agree with lab bhattacharjee that avoiding methods that require later working out signs is a good idea.

Here is a side issue, which in the original version of my answer was all I posted:
$\sqrt{x^2}=|x|$ is an identiy for real numbers $x$.  The reason is that for a nonegative real number $a$, $\sqrt{a}$ is defined to be the unique nonnegative squareroot of $a$.  Since $|x|^2=x^2$, it follows that $|x|$ is the nonnegative real number whose square is $x^2$.
Therefore $\sqrt{\left(\dfrac{\sin x}{1-\cos x}\right)^2}=\left|\dfrac{\sin x}{1-\cos x}\right|$.
