I had to calculate $\tan(π/16)+\tan(5π/16)+\tan(9π/16)+\tan(13π/16)$
I tried writing $π/16=x$ and then writing the sum as $\tan x+\tan(9x-4x)+\tan9x+\tan(9x+4x)$ and then simplifying using $\tan(x+y)$ and $\tan(x-y)$ but it didn't simplify to anything, can anybody please give a hint on how should I approach this.
$$a=\tan(\frac{\pi}{16})+\tan(\frac{5\pi}{16})+\tan(\frac{9\pi}{16})+\tan(\frac{13\pi}{16})\\= \tan(\frac{\pi}{16})+\tan(\frac{5\pi}{16})+\tan(\frac{1\pi}{16}+\frac{\pi}{2})+\tan(\frac{5\pi}{16}+\frac{\pi}{2})\\=\\ \tan(\frac{\pi}{16})-\cot(\frac{\pi}{16})+\tan(\frac{5\pi}{16})-\cot(\frac{5\pi}{16})\\=?$$ now let me note that $\color{red} {2\cot(2x)=\cot x- \tan x}\tag{1}$ so $$a=-(-\tan(\frac{\pi}{16})+\cot(\frac{\pi}{16})-\tan(\frac{5\pi}{16})+\cot(\frac{5\pi}{16}))\\= -(2\cot(\frac{2\pi}{16})+2\cot(\frac{10\pi}{16}))\\=$$can you go on ? $$a=-2(\cot(\frac{\pi}{8})+\cot(\frac{5\pi}{8}))\\=-2(\cot(\frac{\pi}{8})+\cot(\frac{\pi}{8}+\frac{\pi}{2}))\\\text{from formula (1)}\\ -2(\cot(\frac{\pi}{8})-\tan(\frac{\pi}{8}))\\=-2(2\cot(2(\frac{\pi}{8}))\\=-4$$
Hint: $$\tan\left(x+\pi\right) = \tan(x).$$ If $x=\frac{3\pi}{16}$, and $y=\frac{2\pi}{16}$, then $x+y = ?$ and $x-y=?$.
