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
$$=-\sqrt{ \frac{ 1+\sqrt{\frac{1+\cos(5\pi/3)}{2}}}{1-\sqrt{\frac{1+\cos(5\pi/3)}{2}}} }$$

Apart from a couple of threes that might be fours ? You are fine upto here
\begin{eqnarray*}
\tan(-\frac{5\pi}{16})=-\sqrt{ \frac{ 1+\sqrt{\frac{1+\cos(5\pi/4)}{2}}}{1-\sqrt{\frac{1+\cos(5\pi/4)}{2}}}}. 
\end{eqnarray*}
Now $\cos(5\pi/4)=\frac{ 1}{\sqrt{2}}$ & after a little bit algebra ...
\begin{eqnarray*}
\tan(-\frac{5\pi}{16})=-\sqrt{ \frac{ 8^{1/4}+\sqrt{\sqrt{2}-1}}{ 8^{1/4}-\sqrt{\sqrt{2}-1}}}. 
\end{eqnarray*}
My Casio fx-83MS gives both sides equal to $-1.4966 \cdots $.
