\(\tan(\alpha + \beta) = \frac{\tan\alpha + \tan\beta}{1 - \tan\alpha\tan\beta}\)

Numerator: \(\frac{m}{m+1} + \frac{1}{2m+1} = \frac{m(2m+1) + (m+1)}{(m+1)(2m+1)} = \frac{2m^2 + 2m + 1}{(m+1)(2m+1)}\)

Denominator: \(1 - \frac{m}{(m+1)(2m+1)} = \frac{2m^2 + 3m + 1 - m}{(m+1)(2m+1)} = \frac{2m^2 + 2m + 1}{(m+1)(2m+1)}\)

So \(\tan(\alpha + \beta) = 1\), and \(\alpha + \beta = \frac{\pi}{4}\).

Answer: (B) \(\frac{\pi}{4}\)