Summation of tan

Question

What is $$\sum _1 ^7 \tan^2 (\frac {n\pi}{16}) -1$$ so I used $\tan(x)=\cot(\frac {\pi}{2}-x)$ for last three angles ie $5\pi/16,6\pi/16,7\pi/16$ . Thus it gets converted to some symmetry of tan,cot but after that I am not sure how to proceed or which identity to use.Use of AM-GM seems to give a very rough estimate . Answer given is $34$

Answer 1

Use this identity (details can be found here Prove that $\sum\limits_{k=1}^{n-1}\tan^{2}\frac{k \pi}{2n} = \frac{(n-1)(2n-1)}{3}$) for $N=8$: $$\sum_{n=1}^{N-1}\tan^{2}\left(\frac{n \pi}{2N}\right)= \frac{(N-1)(2N-1)}{3}.$$

Answer 2

$$S=\sum^{7}_{n=1}\tan^2 \left(\frac{n\pi}{16}\right) = \tan^2 \left(\frac{\pi}{16}\right)+\tan^2 \left(\frac{\pi}{8}\right)+\tan^2 \left(\frac{3\pi}{16}\right)+1+\tan^2 \left(\frac{5\pi}{16}\right)+\tan^2 \left(\frac{3\pi}{8}\right)+\tan^2 \left(\frac{7\pi}{16}\right)$$

And using $\displaystyle \tan^2 \frac{7\pi}{16} = \cot^2 \frac{\pi}{16}$ and $\displaystyle \tan^2 \frac{5\pi}{16} = \cot^2 \frac{3\pi}{16}$

Using $\displaystyle \tan^2 x+\cot^2 x = \frac{\sin^4 x+\cos^4 x}{\sin^2 x \cos ^2 x} = \frac{4-2(\sin 2x)^2}{\sin ^2 (2x)} = \frac{8-2(1-\cos 4x)}{1-\cos 4x}=\frac{6+2\cos 4x}{1-\cos 4x}$ And Using $\displaystyle \tan^2 \alpha = \frac{2}{\cos 2 \alpha + 1}-1$

So we get $$S = 14+8\sqrt{2}+14-8\sqrt{2}+3-\sqrt{2}+3+2\sqrt{2}+1 = 35$$