As $$\tan(2n+1)s=\frac{t^{2n+1}-\binom{2n+1}2t^{2n-1}+\cdots}{\binom{2n+1}1t^{2n}-\binom{2n+1}3t^{2n-2}+\cdots}$$ where $t=\tan s$
So, $$\tan 7s=\frac{t^7-21t^5+35t^3-7t}{7t^6-35t^4+21t-1}$$
If we put $7s=\pi,t^7-21t^5+35t^3-7t=0--->(1)$ whose roots are $\tan\frac{r\pi}7$ where $r=0,1,2,3,4,5,6$
So, the roots of $t^6-21t^4+35t^2-7=0--->(2)$ are $\tan\frac{r\pi}7$ where $r=1,2,3,4,5,6$
If we put $z=\frac1t$ (as $t\ne0,$) $\frac1{z^6}-\frac{21}{z^4}+\frac{35}{z^2}-7=0\implies z^6-5z^4+3z^2-\frac17=0$ whose roots are $\cot\frac{r\pi}7$ where $r=1,2,3,4,5,6$
So, $$z^6-5z^4+3z^2-\frac17=\prod_{1\le r\le 6}(z-\cot\frac{r\pi}7)$$
But as $\cot\frac{(7-r)\pi}7=\cot(\pi-\frac{r\pi}7)=-\cot\frac{r\pi}7$,
so $\prod_{1\le r\le 6}(z-\cot\frac{r\pi}7)$
$=\prod_{1\le r\le 3}(z-\cot\frac{r\pi}7)\prod_{4\le r\le 6}(z-\cot\frac{r\pi}7)$
$=\prod_{1\le r\le 3}(z-\cot\frac{r\pi}7)\prod_{3\ge u\ge 1}(z+\cot\frac{u\pi}7)$ (putting $7-r=u$)
$=\prod_{1\le r\le 3}(z^2-\cot^2\frac{r\pi}7)$
So,$z^6-5z^4+3z^2-\frac17$
$=z^6-z^4\sum_{1\le r\le 3}\cot^2\frac{r\pi}7$
$+z^2(\cot^2\frac{\pi}7\cot^2\frac{2\pi}7+\cot^2\frac{\pi}7\cot^2\frac{2\pi}7+\cot^2\frac{2\pi}7\cot^2\frac{3\pi}7)-\prod_{1\le r\le 3}\cot^2\frac{r\pi}7$
Comparing the coefficients of $z^4,$ we get the required identity.
Alternatively,
If we put $\cot^2\frac{n\pi}7=y$ where $n=1,2,3$ or $n=7-1,7-2,7-3$
we get $y=\frac1{t^2}$ (as $\cot\frac{(7-r)\pi}7=-\cot\frac{r\pi}7$)
Replacing $t^2=\frac1y$ in $(2)$ we get $\frac1{y^3}-\frac{21}{y^2}+\frac{35}y-7=0$
or $7y^3-35y^2+21y-1=0$
So, $\sum_{1\le n\le 3}\cot^2\frac{n\pi}7=\frac{35}7=5$ using Vieta's Formulae.