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?