To solve the problem, we start with the given equation:

\(2 \sec 4\beta = \tan 2\alpha + \cot 2\alpha\)

First, observe the identity for the tangent and cotangent:

\(\tan 2\alpha + \cot 2\alpha = \frac{\sin 2\alpha}{\cos 2\alpha} + \frac{\cos 2\alpha}{\sin 2\alpha}\)

Combine the terms over a common denominator:

\(\tan 2\alpha + \cot 2\alpha = \frac{\sin^2 2\alpha + \cos^2 2\alpha}{\sin 2\alpha \cos 2\alpha}\)

Using the Pythagorean identity, \(\sin^2 2\alpha + \cos^2 2\alpha = 1\), the expression simplifies to:

\(\tan 2\alpha + \cot 2\alpha = \frac{1}{\sin 2\alpha \cos 2\alpha}\)

Next, substitute \(\sin 2\alpha \cos 2\alpha\) using the double-angle identity:

\(\sin 2\alpha \cos 2\alpha = \frac{1}{2} \sin 4\alpha\)

Thus, the equation becomes:

\(2 \sec 4\beta = \frac{2}{\sin 4\alpha}\)

Or:

\(\sec 4\beta = \frac{1}{\sin 4\alpha}\)

Since \(\sec 4\beta = \frac{1}{\cos 4\beta}\), we equate:

\(\cos 4\beta = \sin 4\alpha\)

One solution to this trigonometric equation is:

\(4\beta = \left(\frac{\pi}{2} - 4\alpha\right) + n\pi\) for some integer \(n\)

Set \(n = 0\) for simplicity:

\(4\beta = \frac{\pi}{2} - 4\alpha\)

Add \(4\alpha\) to both sides:

\(4(\alpha + \beta) = \frac{\pi}{2}\)

Solve for \(\alpha + \beta\):

\(\alpha + \beta = \frac{\pi}{8}\)

Therefore, the correct answer is:

\(\frac{\pi}{8}\)