Finding $\frac{\sum_{r=1}^8 \tan^2(r\pi/17)}{\prod_{r=1}^8 \tan^2(r\pi/17)}$ I have tried to wrap my head around this for some time now, and quite frankly I am stuck.
Given is that :
$$a=\sum_{r=1}^8 \tan^2\left(\frac{r\pi}{17}\right) \qquad\qquad b=\prod_{r=1}^8 \tan^2\left(\frac{r\pi}{17}\right)$$
Then what is the value of $a/b$?
I tried evaluating the quantities on DESMOS to realise that: a=136 and b=17 which gives 8 as the answer. Any helpful insight about how to reach at these values !?
 A: Here is a solution using the observation in Cauchy's proof of Basel problem.

Lemma. We have
$$ \prod_{k=1}^{n} \left( t - \tan^2\left(\frac{k\pi}{2n+1}\right)\right) = \sum_{k=0}^{n} (-1)^{n-k}\binom{2n+1}{2k+1} t^k. \tag{*} $$

Using this, we immediately know that
$$ a = \sum_{k=1}^{n} \tan^2\left(\frac{k\pi}{2n+1}\right) = \binom{2n+1}{2} = n(2n+1) $$
and
$$ b = \prod_{k=1}^{n} \tan^2\left(\frac{k\pi}{2n+1}\right) = \binom{2n+1}{1} = 2n+1. $$
So the ratio $a/b$ is exactly $n$. In OP's case, $n = 8$ and hence the answer is $8$.

Proof of Lemma. Write $m = 2n+1$. Then by the de Moivre's formula,
\begin{align*}
\frac{\cos(mx) + i\sin(mx)}{\cos^m x} = \left( \frac{\cos x + i\sin x}{\cos x} \right)^m = (1 + i\tan x)^m = \sum_{k=0}^{m} \binom{m}{k} (i\tan x)^k.
\end{align*}
Taking the imaginary parts only and plugging $m = 2n+1$ back,
\begin{align*}
\frac{\sin((2n+1)x)}{\cos^{2n+1} x} = \sum_{k=0}^{n} (-1)^k \binom{m}{2k+1} \tan^{2k+1} x,
\end{align*}
and so,
\begin{align*}
(-1)^n \frac{\sin((2n+1)x)}{\sin x \cos^{2n} x} = \sum_{k=0}^{n} (-1)^{n-k} \binom{m}{2k+1} \tan^{2k} x.
\end{align*}
However, the left-hand side becomes zero if $x_k = \frac{k\pi}{2n+1}$ for $k = 1, \dots, n$, and also note that $\tan^2(x_k)$ for $k = 1, \dots, n$ are all different. So, denoting the right-hand side of $\text{(*)}$ by $P(t)$, i.e., writing
$$ P(t) = \sum_{k=0}^{n} (-1)^{n-k}\binom{2n+1}{2k+1} t^k, $$
then this implies that $P(\tan^2(x_k)) = 0$ for $k = 1, \dots, n$, hence $\tan^2(x_k)$ for $k = 1, \dots, n$ are $n$ distinct zeros of $P(t)$. Since the leading coefficient of $P(t)$ is $1$, this proves $\text{(*)}$.
A: Like Trig sum: $\tan ^21^\circ+\tan ^22^\circ+\cdots+\tan^2 89^\circ = \text{?}$,
$$\tan(2n+1)x=\dfrac{\binom{2n+1}1t-\binom{2n+1}1t^3+\cdots+(-1)^n\binom{2n+1}{2n+1}t^{2n+1}}{\binom{2n+1}0-\binom{2n+1}2t^2+\cdots+(-1)^n\binom{2n+1}{2n}t^{2n}}$$
where $t=\tan x$
So, the roots of $$\binom{2n+1}1t-\binom{2n+1}1t^3+\cdots+(-1)^n\binom{2n+1}{2n+1}t^{2n+1}=0$$  i.e., $$t^{2n+1}-\binom{2n+1}{2n-1}t^{2n-1}+\cdots+(-1)^n(2n+1)t=0$$ are $t_r=\tan\dfrac{r\pi}{2n+1}; r=0,\pm1,\pm2,\cdots,\pm n$
As $r=0\implies \tan\dfrac{r\pi}{2n+1}=0$, the roots of $$t^{2n}-\binom{2n+1}{2n-1}t^{2n-2}+\cdots+(-1)^n(2n+1)=0$$  are $t_r=\tan\dfrac{r\pi}{2n+1}; r=\pm1,\pm2,\cdots,\pm n$
Writing $t_r^2=p_r,$
the roots of $$p^n-\binom{2n+1}{2n-1}p^{n-1}+\cdots+(-1)^n(2n+1)=0$$  are $p_r=\tan^2\dfrac{r\pi}{2n+1}; r=1,2,\cdots, n$
Use Vieta's formula
A: $$=∑_{=1}^8 tan^2(\frac{}{17})=136$$
$$b=\prod_{=1}^8 tan^2(\frac{}{17})= 17$$
$$\frac{a}{b}=\frac{136}{17}=8$$
This results can be calculated by hand.
$${Tan[\frac{}{17}]^2, Tan[(2\frac{}{17}]^2, Tan[(3 \frac{}{17}]^2, 
 Tan[(4 \frac{}{17}]^2, Cot[(7 \frac{}{34}]^2, Cot[(5 \frac{}{34}]^2, 
 Cot[(3 \frac{}{34}]^2, Cot[\frac{}{34}]^2}$$
$${0.0349437, 0.15008, 0.383376, 0.831052, 1.75354, 4.03315, 12.3527, \
116.461}$$
This confirms the formulas by @sangchul-lee: Cauchys proof of the Basel lemma but with the figures calculated.
