Evaluate $(\sum_{k=1}^{7} \tan^2(\frac{k\pi}{16})) - \tan^2(\frac{4\pi}{16})$ The question is:

Evaluate 
  $$\displaystyle \left(\sum_{k=1}^{7} \tan^2\left(\frac{k\pi}{16}\right)\right) - \left(\tan^2\frac{4\pi}{16}\right)$$
  The given answer:$34$


What I've tried:
$$\displaystyle \left(\sum_{k=1}^{7} \tan^2\left(\frac{k\pi}{16}\right)\right) - \left(\tan^2\frac{4\pi}{16}\right)$$
$$=\tan^2\left(\frac{\pi}{16}\right)+\tan^2\left(\frac{2\pi}{16}\right)+\tan^2\left(\frac{3\pi}{16}\right)+\tan^2\left(\frac{5\pi}{16}\right)+\tan^2\left(\frac{6\pi}{16}\right)+\tan^2\left(\frac{7\pi}{16}\right)$$
$$=\tan^2\left(\frac{\pi}{16}\right)+\tan^2\left(\frac{2\pi}{16}\right)+...+\tan^2\left(\frac{\pi}{2}-\frac{2\pi}{16}\right)+\tan^2\left(\frac{\pi}{2}-\frac{\pi}{16}\right)$$
$$=\tan^2\left(\frac{\pi}{16}\right)+\cot^2\left(\frac{\pi}{16}\right)+\tan^2\left(\frac{2\pi}{16}\right)+\cot^2\left(\frac{2\pi}{16}\right)+\tan^2\left(\frac{3\pi}{16}\right)+\cot^2\left(\frac{3\pi}{16}\right)$$

How do I proceed from here?
 A: Using $\displaystyle \tan^2(x)+\cot^2(x)=\frac{\sin^4(x)+\cos^4(x)}{\sin^2(x)\cdot \cos^2(x)}=\frac{1-2\sin^2 x\cos^2 x}{\sin^2 x\cos^2 x}.$
So we have $\displaystyle =4\csc^2(2x)-2=2+4\cot^2(2x).$
So our expression is
$\displaystyle 2+4\cot^2(\pi/8)+2+4\cot^2(\pi/4)+2+4\cot^2(3\pi/8)$
$\displaystyle 6+4+4\bigg[\cot^2(\pi/8)+\tan^2(\pi/8)\bigg]$
$\displaystyle =10+4\bigg[2+4\cot^2(\pi/4)\bigg]$
$=10 + 4\bigg[2+4\bigg]=34.$
A: 
The summation is 35 and subtract -1 because tan pi/4 is 1
Finally the result is 34
A: Using Expansion of $\tan(nx)$ in powers of $\tan(x)$,
$\tan\dfrac{k\pi}{2n},0\le k<2n,k\ne n$ are the roots of
$$\binom{2n}1t-\binom{2n}3t^3+\cdots+(-1)^{n-1}\binom{2n}{2n-1}t^{2n-1}=0$$
$$\implies \binom{2n}{2n-1}t^{2n-2}-\binom{2n}{2n-3}t^{2n-3}+\cdots+(-1)^{n-1}\binom{2n}1=0$$ will have the roots $\tan\dfrac{k\pi}n, 1\le k<2n;k\ne n$
So, the roots of $$2nc^{n-1}-\binom{2n}3c^{n-2}+\cdots=0$$ will be $c_k=\tan^2\dfrac{k\pi}n,1\le k<n$
$$\sum_{k=1}^{n-1}c_k=\dfrac{\binom{2n}3}{2n}=?$$
Here $2n=16$
