In old popular science magazine for school students I've seen problem
Prove that $\quad $ $\dfrac{1}{\cos^2 20^\circ} + \dfrac{1}{\cos^2 40^\circ} + \dfrac{1}{\cos^2 60^\circ} + \dfrac{1}{\cos^2 80^\circ} = 40. $
How to prove more general identity:
$$ \begin{array}{|c|} \hline \\ \sum\limits_{k=0}^{n-1}\dfrac{1}{\cos^2\frac{\pi k}{n}}=n^2 \\ \hline \end{array} , \qquad \mbox{ where } \ n \ \mbox{ is odd.}$$
Whenever I see a problem like this, I think Chebyshev polynomials.
The Chebyshev polynomials of the first kind, $T_m(x)$, are defined so that:
$$T_m(\cos \theta) = \cos m\theta$$
$T_m$ is an $m$th degree polynomial, and the roots of $T_m(x)-1$ are exactly $\cos 2\pi k/m$ for $k=0,1,\dots,m-1$.
When $m=2n$ is even, we can write $$T_{2n}(x)=1+C\prod_{k=0}^{n-1}\left(x^2-\cos^2\frac{k\pi}{n}\right)$$ for some constant $C$. The occurrence of those $\cos^2\frac{k\pi}{n}$ suggested this might be a good approach.
Now, given a polynomial $p(x)=a_mx^m + a_{m-1}x^{m-1}\dots +a_0$, with non-zero roots $r_0,\dots, r_{m-1}$, we have formulas: $$\sum \frac{1}{r_k} = -\frac{a_1}{a_0}$$ And: $$\sum_{i< j} \frac{1}{r_ir_j} = \frac{a_2}{a_0}$$
So $$\sum \frac{1}{r_k^2} = \left(-\frac{a_1}{a_0}\right)^2 - 2\frac{a_2}{a_0}$$
Now let $p(x)=T_{2n}(x)-1$. The roots of $p$ are $r_k=\cos \frac{k\pi}{n}$ for $k=0,\dots,2n-1$, and the roots are non-zero since $n$ is odd. We know that $p(x)$ is even, so $a_1=0$. Finally, we know that $\sum_{k=0}^{2n-1} \frac{1}{r_k}^2$ is twice the sum that you are looking for.
So your sum is now reduced to finding $-\frac{a_2}{a_0}$ where the $a_0,a_2$ are coefficients of $p(x)$. Since $p(0)=T_{2n}(\cos \pi/2)-1 = \cos n\pi - 1 = -2$, so we know that $a_0=-2$. So $\sum r_k^{-2} = a_2$. (Again, we use $n$ odd here.)
So we need to prove that $a_2=2n^2$.
Now, $a_2=\frac{1}{2}T_{2n}^{''}(0)$. Let $f(x)=\cos 2nx = T_{2n}(\cos x)$. Differentiating we get:
$$-2n\sin 2nx = -T_{2n}^{'}(\cos x)\sin x$$
Differentiating both sides again:
$$-4n^2\cos 2nx = T_{2n}^{''}(\cos x)(\sin^2 x) - T_{2n}^{'}(\cos x)\cos x$$
Putting in $x=\pi/2$, then $\cos x=0$, $\sin x=1$, and $\cos 2nx=-1$. Therefore, we get $$4n^2 = T_n^{''}(0)\\a_n=\frac{1}{2}T_{2n}^{''}(0)=2n^2$$
and therefore your sum is $n^2$.
Note: Any symmetric rational function with rational coefficients of $\{\cos 2\pi k/n\mid k=0,\dots,n-1\}$ will be rational by this argument.
Since $n$ is odd, the numbers $u_k=\cos k\pi/n$ are the same as the numbers $\cos 2k\pi/n$, i.e. the distinct angles $\theta$ satisfying $n\theta=0$ (mod $2\pi$). We think of them as the roots of the equation $\cos n\theta=1$. Writing $$\cos n\theta = \cos^n\theta - \binom{n}{2}\cos^{n-2}\theta\sin^2\theta \cdots \pm n\cos\theta \sin^{n-1}\theta$$ and using $\sin^2\theta=1-\cos^2\theta$, we see that the $u_k$'s are the roots of the polynomial $$p(u)=u^n - \binom{n}{2}u^{n-2}(1-u^2) \pm n u(1-u^2)^{(n-1)/2} + 1.$$ Note that all powers of $u$ which occur are odd (except for the constant term). The reciprocals $1/u_k$ are the roots of the "reverse polynomial" $$r(u)=u^n p(1/u) = u^n + a_{n-1} u^{n-1} + \cdots,$$ where $a_{n-1}=\pm n$ and $a_{n-2}=0$.
The sum in question is the sum of the squares of the roots of $r(u)$, i.e. $a_{n-1}^2 - 2a_{n-2} = n^2$.
I believe I first came across this trick in this paper by Szenes, so check out section 3 in that for more details.
Consider the form $$\mu_n(z) = n\frac{dz}{z} \frac{z^n + z^{-n}}{z^n-z^{-n}}$$ and the function $$f(z) = \frac{4}{(z+z^{-1})^2}.$$ Your sum is the sum of the $f(z_k)$, where $z_k = \exp(\pi i k/n)$, $k = 0, \dots, n-1$. Note first of all that for these particular values of $z$, $$f(z_k) = \mathop{\mathrm{Res}}_{z=z_k} f \mu_n.$$ Note also that the same formula holds with $z_k$ replaced by $z_k^{-1}$, and that $\mathop{\mathrm{Res}}_{z=\infty} f\mu_n = \mathop{\mathrm{Res}}_{z=0} f\mu_n = 0.$ Moreover, $\mathop{\mathrm{Res}}_{z= \pm i} f\mu_n = -n^2$. It follows from the residue theorem that $$\sum_{k=0}^{n-1} f(z_k) = -\frac{1}{2}\left(\mathop{\mathrm{Res}}_{z=i} f\mu_n + \mathop{\mathrm{Res}}_{z=-i} f\mu_n\right) = n^2.$$
In equation $(7)$ of this answer, I compute that $$ \sum_{k=1}^n\tan^2\left(\frac{k\pi}{2n+1}\right)=n(2n+1) $$ Thus, with $n=2m+1$, $$ \begin{align} \sum_{k=0}^{n-1}\frac1{\cos^2\left(\frac{k\pi}{n}\right)} &=\sum_{k=0}^{2m}\sec^2\left(\frac{k\pi}{2m+1}\right)\\ &=2\sum_{k=0}^m\tan^2\left(\frac{k\pi}{2m+1}\right)+2(m+1)\\[6pt] &=2m(2m+1)+2(m+1)\\[12pt] &=(2m+1)^2\\[12pt] &=n^2 \end{align} $$
Residue Theory Answer
For odd $n$, consider the function
$$
f(z)=\frac{n/z}{z^n-1}\left(\frac{z-1}{z+1}\right)^2\tag{1}
$$
All the singularities are simple, except the singularity at $-1$.
$$
\left(\frac{z-1}{z+1}\right)^2=1-\frac4{z+1}+\frac4{(z+1)^2}\tag{2}
$$
Furthermore,
$$
\frac{\mathrm{d}}{\mathrm{d}z}\frac{n/z}{z^{\raise{2pt}n}-1}=\frac{n-n(n+1)z^n}{z^2(z^{\raise{2pt}n}-1)^2}\tag{3}
$$
Using $(2)$ and $(3)$, we get the residue of $f$ at $-1$ to be
$$
-4\cdot\frac{-n}{-1-1}+4\cdot\frac{n+n(n+1)}{(-1-1)^2}=n^2\tag{4}
$$
The residue of $f$ at $0$ is $-n$ and the residue at each $n^{\text{th}}$ root of unity is $-\tan^2(\theta/2)$.
Since the integral of $f$ over an increasing circle vanishes, the sum of the residues must be $0$. Therefore,
$$
\sum_{k=0}^{n-1}\tan^2\left(\frac{k\pi}{n}\right)=n^2-n\tag{5}
$$
and therefore,
$$
\sum_{k=0}^{n-1}\sec^2\left(\frac{k\pi}{n}\right)=n^2\tag{6}
$$
From this answer we know that $$ \sum\limits_{k=1}^{m}\tan^2\frac{\pi k}{2m+1}=m(2m+1) $$ Similarly $$ \sum\limits_{k=m+1}^{2m}\tan^2\frac{\pi k}{2m+1}= \sum\limits_{l=1}^{m}\tan^2\frac{\pi (2m+1-l)}{2m+1}= \sum\limits_{l=1}^{m}\tan^2\frac{\pi l}{2m+1}=m(2m+1) $$ Since $\cos^{-2}\alpha=1+\tan^2\alpha$, then $$ \begin{align} \sum\limits_{k=0}^{2m}\frac{1}{\cos^2\frac{\pi k}{2m+1}} &=\sum\limits_{k=0}^{2m} 1 +\sum\limits_{k=0}^{2m}\tan^2\frac{\pi k}{2m+1}\\ &=2m+1+\sum\limits_{k=1}^{m}\tan^2\frac{\pi k}{2m+1}+\sum\limits_{k=m+1}^{2m}\tan^2\frac{\pi k}{2m+1}\\ &=2m+1+m(2m+1)+m(2m+1)=(2m+1)^2 \end{align} $$
