Prove that $\frac{1}{4-\sec^{2}(2\pi/7)} + \frac{1}{4-\sec^{2}(4\pi/7)} + \frac{1}{4-\sec^{2}(6\pi/7)} = 1$ 
How can I prove the fact $$\frac{1}{4-\sec^{2}\frac{2\pi}{7}} + \frac{1}{4-\sec^{2}\frac{4\pi}{7}} + \frac{1}{4-\sec^{2}\frac{6\pi}{7}} = 1.$$

When asked somebody told me to use the ideas of Chebyshev polynomial, but I haven't learnt that in school. 
I tried doing this way:
Look at $y =\cos\theta + i \sin\theta$ where $\displaystyle\theta \in \Bigl\{\frac{2\pi}{7},\frac{4\pi}{7},\cdots,2\pi\Bigr\}$
Then we have 
\begin{align*}
y^{7} &=1 \\ y^{7}-1 &=0 \\ (y-1) \cdot (y^{6}+y^{5}+\cdots + 1) &= 0
\end{align*}
Now the root $y=1$ corresponds to $\theta = 2\pi$, and that $$y^{6} + y^{5}+\cdots + 1 =0$$
have roots $\cos\theta + i \sin\theta$, where $\theta \in \Bigl\{\frac{2\pi}{7},\frac{4\pi}{7} ,\cdots \Bigr\}$. Looking at $y+\frac{1}{y} $ will give me the roots as $\cos\theta$ and then i can put $z=y^{2}$ to get $\cos^{2}$ as the roots and the invert to get $\sec^{2}$, but I have some problems. 
Can anyone help me out with a neat solution. 
Thanks.
 A: Where you have left of $y^6+y^5+\cdots+y+1=0$ where $y=\cos \theta+i\sin \theta$ where $\theta=\frac{2\pi}7,\frac{4\pi}7,\frac{6\pi}7,\cdots , \frac{12\pi}7$
Let us divide both sides by $y^3,$ $y^3+\frac1{y^3}+y^2+\frac 1{y^2}+y+\frac 1 y+1=0$
or $\left(y+\frac1y\right)^3-3\left(y+\frac1y\right)+\left(y+\frac1y\right)^2-2+\left(y+\frac1y\right)+1=0$
or $\left(y+\frac1y\right)^3+\left(y+\frac1y\right)^2-3\left(y+\frac1y\right)-1=0$
Now,  $\displaystyle y+\frac 1 y=2\cos \theta=z$ (say)
So, $\displaystyle z^3+z^2-3z-1=0\ \ \ \ \color{Red}{(1)},$ has the roots $\displaystyle 2\cos\frac{2\pi}7, 2\cos\frac{4\pi}7, 2\cos\frac{6\pi}7$ using $\displaystyle\cos\frac{r\pi}7=\cos\left(2\pi-\frac{r\pi}7\right)=\cos\frac{(14-r)\pi}7 $ as $\color{Red}{(1)}$ does not have repeated roots
$\displaystyle\implies z^2(1+z)=3z-1,z^2=\frac{3z-1}{z+1}$
$$\text{Now,}\displaystyle\frac 1{4-\sec^2\theta}=\frac{\cos^2\theta}{4\cos^2\theta-1}= \frac{z^2}{4z^2-4}=w(say),$$ 
$\displaystyle\implies z^2=\frac {4w}{4w-1}$
Comparing the values of $\displaystyle z^2, \frac {4w}{4w-1}=\frac{3z-1}{z+1}$
Replacing the $z$ with $w$ in $\color{Red}{(1)}$, we shall get a cubic equation in $w,$  whose sum of roots will give us the required identity.
A: 
Motto: Use as much as possible the set of complex roots of unity as a whole rather than each complex root in isolation. 
Sub-motto: The complex exponential is easier to use than the sine and cosine functions hence reducing the latter ones to the former is often fruitful.

Note that, for every $x$ such that this fraction exists, 
$$
\frac1{4-\sec^2x}=\frac14+\frac14\frac1{2\cos(2x)+1}=\frac14+\frac14\frac{\mathrm e^{2\mathrm ix}}{\mathrm e^{4\mathrm ix}+\mathrm e^{2\mathrm ix}+1},
$$ 
hence the identity to be proved is 
$$
\sum\limits_z\frac{z^2}{z^4+z^{2}+1}=1,
$$ 
where the sum is over $z$ in $\{\omega,\omega^2,\omega^3\}$ with $\omega=\mathrm e^{2\mathrm i\pi/7}$.
If $z^7=1$ and $z\ne1$, then $\dfrac{z^2}{z^4+z^2+1}=-z^3-z^{4}$
hence the sum in the LHS is
$$
-\omega^3-\omega^{4}-\omega^6-\omega^{8}-\omega^9-\omega^{12},
$$ 
that is,
$$
-\omega^3-\omega^{4}-\omega^6-\omega-\omega^2-\omega^5=1-\sum\limits_{k=1}^7\omega^k=1.
$$
