$$ \mbox{Question: Evaluate}\quad \tan^{2}\left(\pi \over 16\right) + \tan^{2}\left(2\pi \over 16\right) + \tan^{2}\left(3\pi \over 16\right) + \cdots + \tan^{2}\left(7\pi \over 16\right) $$
What I did: Well I know that $\tan^{2}\left(7\pi/16\right)$ is the same as $\cot^{2}\left(\pi/16\right)$. Thus this will repeat for all values up to $\tan^{2}\left(4\pi/16\right)$.
However, I don't understand where to proceed from there.
The 15 numbers $\tan k\pi/16$, $-7\leq k\leq7$, are the roots of the polynomial (of degree 15) $(1+ix)^{16}-(1-ix)^{16}$ (using the geometric interpretation of complex numbers and their multiplication). The polynomial is of the form $-32ixp(x^2)$, where $p(t)=t^7-35t^6+\dots$ is a polynomial of degree 7, the roots of $p$ are thus $\tan^2 k\pi/2$, $1\leq k\leq7$. Your sum is therefore $35$.
HINT: Use the half angle formula with $\theta = \pi/4$ to find $\tan(\pi/8)$ and do the same with $\theta = \pi/8$ to find $\tan(\pi/16)$
EDIT: The half angle formula is: $$\tan(a) = \frac{2\tan(\frac{a}{2})}{1- \tan^2(\frac{a}{2})}$$ So use this formula for $a = \pi/4$ and $a = \pi/8$
Like Find $\sum\limits_{k=1}^{12}\tan \frac{k\pi}{13}\cdot \tan \frac{3k\pi}{13}$,
$$\tan16\theta=\frac{\binom{16}1t-\binom{16}3t^3+\cdots+\binom{16}{13}t^{13}-\binom{16}{15}t^{15}}{\cdots}\text{ where }t=\tan\theta$$
If $\displaystyle\tan16\theta=0,16\theta=r\pi$ where $r$ is any integer
$\displaystyle\implies\theta=\frac{r\pi}{16}$ where $0\le r\le15$
So, the roots of $\displaystyle\binom{16}1t-\binom{16}3t^3+\cdots+\binom{16}{13}t^{13}-\binom{16}{15}t^{15}=0$ are $\displaystyle\tan r\theta$ where $\displaystyle0\le r\le15,r\ne8$ as $\displaystyle\tan\frac{8\cdot\pi}{16}$ is not finitely defined
So, the roots of $\displaystyle\binom{16}1-\binom{16}3t^2+\cdots+\binom{16}{13}t^{12}-\binom{16}{15}t^{14}=0$
or of $\displaystyle\binom{16}{15}t^{14}-\binom{16}{13}t^{12}+\cdots+\binom{16}3t^2-\binom{16}1=0$
are $\displaystyle\tan r\theta$ where $\displaystyle1\le r\le15,r\ne8$
But as $\displaystyle\tan\frac{(16-n)\pi}{16}=\tan\left(\pi-\frac{n\pi}{16}\right)=-\tan\frac{n\pi}{16},\tan^2\frac{(16-n)\pi}{16}=\tan^2\frac{n\pi}{16};$
the roots of $\displaystyle\binom{16}{15}u^7-\binom{16}{13}u^6+\cdots+\binom{16}3u-\binom{16}1=0$ are $\displaystyle \tan^2\frac{(16-n)\pi}{16}=\tan^2\frac{n\pi}{16}$ where $1\le n\le 7$ or $9\le r\le15$
Now Vieta's formula is inviting
