How to find $\tan(-\frac{5\pi}{16})$ with half-angle formulas? How to find $\tan(-\frac{5\pi}{16})$ with half-angle formulas?
I tried the $\pm \sqrt{\frac{1-\cos{A}}{1+\cos{A}}}$ and $\frac{\sin{A}}{1+\cos{A}}$ but got stuck once there were square roots on top and bottom like $\frac{\sqrt{...}}{1-\sqrt{...}}.$
Using the cosine over cosine in square root I got up to
$$=-\sqrt{ \frac{ 1+\sqrt{\frac{1+\cos(5\pi/3)}{2}}}{1-\sqrt{\frac{1+\cos(5\pi/3)}{2}}} }$$

The problem is straightforward, it's just a form which is more trouble to rationalize that it's worth. And there are several equivalent expressions depending upon which identities are used.
$$ \tan\left(-\frac{5\pi}{16}\right)=-\frac{\sin\left(\frac{5\pi}{8}\right)}{1+\cos\left(\frac{5\pi}{8}\right)}$$
and 
$$ \sin\left(\frac{5\pi}{8}\right)=\sqrt{\frac{1-\cos(5\pi/4)}{2}}=\frac{1}{2}\sqrt{2+\sqrt{2}}$$
whereas
$$ \cos\left(\frac{5\pi}{8}\right)=-\sqrt{\frac{1+\cos(5\pi/4)}{2}}=-\frac{1}{2}\sqrt{2-\sqrt{2}}$$
Therefore
$$\tan\left(-\frac{5\pi}{16}\right)=\frac{\sqrt{2+\sqrt{2}}}{\sqrt{2-\sqrt{2}}-2} $$
It takes many steps to rationalize this denominator so one may as well leave it in this form.
