Explanation

(b)
We have, $\tan{\alpha = \frac{m}{m + 1}and{\ tan}{\beta = \frac{1}{2m + 1}}}$
We know that$,\tan(\alpha + \beta) = \frac{\tan{\alpha + \tan\beta}}{1 - \tan{\alpha\tan\beta}}$
$= \frac{\frac{m}{m + 1} + \frac{1}{2m + 1}}{1 - \frac{m}{(m + 1)}.\frac{1}{(2m + 1)}}$
$= \frac{2m^{2} + m + m + 1}{2m^{2} + m + 2m + 1 - m}$
$= \frac{2m^{2} + 2m + 1}{2m^{2} + 2m + 1} = 1$
$\Rightarrow \tan(\alpha + \beta) = \tan\frac{\pi}{4}$
$\Rightarrow \alpha + \beta = \frac{\pi}{4}$­