Evaluate $\sec \frac{2\pi}{7}+\sec \frac{4\pi}{7}+\sec \frac{6\pi}{7}$ The problem is as follows:
Find the value of $\textrm{H}$ which belongs to a certain vibration coming from a magnet.
$$H=\sec \frac{2\pi}{7}+\sec \frac{4\pi}{7}+\sec \frac{6\pi}{7}$$
It was easy to spot that each term was related to multiples of two and three of the first angle. So I rewrote that equation like this:
$$H=\sec \frac{2\pi}{7}+\sec \frac{2\times 2\pi}{7}+\sec \frac{3\times 2\pi}{7}$$
One method which I tried was to transform the multiples of each angle into their equivalents as a single one as shown below:
$$\cos^{2}\omega=\frac{1+\cos 2\omega}{2}$$
$$\cos 2\omega= 2 \cos^{2}\omega - 1$$
$$\cos^{3}\omega=\frac{1}{4}\left(3cos\omega+\cos 3\omega \right)$$
$$\cos 3\omega = 4 \cos^{3}\omega - 3 cos\omega$$
Therefore by plugin these expressions into the above equation would become into (provided that secant function is expressed in terms of secant):
$$H=\frac{1}{\cos \frac{2\pi}{7}}+\frac{1}{2\cos^{2}\frac{2\pi}{7}-1}+\frac{1}{4\cos^{3}\frac{2\pi}{7}-3\cos\omega}$$
But from here on it looks convoluted or too algebraic to continue. My second guess was it could be related to sum to product identity but I couldn't find one for the secant.
Does it exist a shortcut or could it be that am I missing something? Can somebody help me to find the answer?
Can this problem be solved without requiring to use Euler's formulas?

Here is the evaluation based on familiar trigonometric identities. Let $\theta = \pi/7$ and express $H$ in terms of cosine functions
$$H=\frac{1}{\cos2\theta} + \frac{1}{\cos4\theta} + \frac{1}{\cos6\theta}$$
or, in the form of common denominator,
$$H =\frac{\cos2\theta + \cos4\theta + \cos6\theta}{\cos2\theta \cos4\theta \cos6\theta}=\frac ND \tag{1}$$
where we used $\cos(x+y)+\cos(x-y)=2\cos x\cos y$ and the relationships $\cos 4\theta = \cos 10\theta$ and $\cos 6\theta = \cos 8\theta$. To compute the denominator, apply  $\sin 2x = 2\sin x \cos x$ to the denominator three times,
$$ D = \frac{\sin 4\theta \cos 4\theta\cos 6\theta}{2\sin 2\theta} = \frac{\sin 8\theta \cos 8\theta }{4\sin 2\theta}= \frac{\sin 16\theta}{8\sin 2\theta} = \frac{1}{8} \tag{2} $$ 
To compute the numerator, write it as
$$ N =\frac{1}{\sin 2\theta} \left({\sin 2\theta\cos2\theta + \sin 2\theta\cos4\theta + \sin 2\theta\cos6\theta} \right)$$
and apply $\sin(x+y)+\sin(x-y)=2\sin x\cos y$ and $\sin 6\theta = -\sin 8\theta$ to get
$$ N = \frac{\sin 4\theta + (\sin 6\theta - \sin 2\theta) + (\sin 8\theta - \sin 4\theta)}{2\sin 2\theta}=-\frac12\tag 3$$
Finally, plugging (2) and (3) into (1), we arrive at
$$H=-4$$
