To solve the problem, we need to analyze the given conditions: the angles \(\alpha\)and \(\beta\) are complementary, which means they satisfy:

• \(\alpha + \beta = \frac{\pi}{2}\)

We are also given that:

• \(\alpha - \beta = \frac{\pi}{6}\)

Now, let's solve these equations. Add the two equations:

• \(\alpha + \beta + \alpha - \beta = \frac{\pi}{2} + \frac{\pi}{6}\)

This simplifies to:

• \(2\alpha = \frac{\pi}{2} + \frac{\pi}{6}\)

Simplifying the right side:

• \(\frac{\pi}{2} = \frac{3\pi}{6}\) and so:
• \(2\alpha = \frac{3\pi}{6} + \frac{\pi}{6} = \frac{4\pi}{6} = \frac{2\pi}{3}\)

Dividing by 2:

• \(\alpha = \frac{\pi}{3}\)

Using \(\alpha + \beta = \frac{\pi}{2}\), substitute for \(\alpha\):

• \(\frac{\pi}{3} + \beta = \frac{\pi}{2}\)

Thus:

• \(\beta = \frac{\pi}{2} - \frac{\pi}{3} = \frac{3\pi}{6} - \frac{2\pi}{6} = \frac{\pi}{6}\)

Now, we know:

• \(\alpha = \frac{\pi}{3}\)
• \(\beta = \frac{\pi}{6}\)

The problem also states that \(m \tan \beta = n \tan \alpha\). So, substitute the known values of \(\alpha\) and \(\beta\):

• \(m \tan\left(\frac{\pi}{6}\right) = n \tan\left(\frac{\pi}{3}\right)\)

Further simplifying, we use the known values:

• \(\tan\left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}}\)
• \(\tan\left(\frac{\pi}{3}\right) = \sqrt{3}\)

Thus, the equation becomes:

• \(m \cdot \frac{1}{\sqrt{3}} = n \cdot \sqrt{3}\)
• \(\frac{m}{\sqrt{3}} = n \sqrt{3}\)

Cross-multiplying gives:

• \(m = n \cdot 3\)

Now we substitute this relation into the expression we need to evaluate:

• \(\frac{m+n}{m-n} = \frac{3n + n}{3n - n} = \frac{4n}{2n} = 2\)

The correct answer is thus 2.