<s>[INST] 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
$$=-\sqrt{ \frac{ 1+\sqrt{\frac{1+\cos(5\pi/3)}{2}}}{1-\sqrt{\frac{1+\cos(5\pi/3)}{2}}} }$$
 [/INST] Fill in details:
$$r:=\tan\left(-\frac{5\pi}{16}\right)=\tan\frac{11\pi}{16}\implies$$
$$s:=\tan\frac{11\pi}8=\tan\left(2\cdot\frac{11\pi}{16}\right)=\frac{2\tan\frac{11\pi}{16}}{1-\tan^2\frac{11\pi}{16}}=\frac{2r}{1-r^2}$$
and
$$-1=\tan\frac{11\pi}4=\tan\left(2\cdot\frac{11\pi}8\right)=\frac{2s}{1-s^2}\implies$$
$$s^2-2s-1=0\implies s_{1,2}=1\pm\sqrt2$$
and now
$$sr^2+2r-s=0\implies r_{1,2}=\frac{-2\pm\sqrt{4+4s^2}}{2s}=\frac{-1\pm\sqrt{1+s^2}}{s}$$
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