$\Rightarrow \tan \beta = \frac{1}{3}$
==> $\tan 2\beta = \frac{{2\tan \beta }}{{1 - {{\tan }^2}\beta }} = \frac{3}{4}$
$\therefore \tan (\alpha + 2\beta ) = \frac{{\frac{1}{7} + \frac{3}{4}}}{{1 - \frac{1}{7}.\frac{3}{4}}} = \frac{{25}}{{25}} = 1$
Now, $0 < \beta < \frac{\pi }{2}$ and $\tan 2\beta = \frac{3}{4} > 0$ both
==> $0 < 2\beta < \frac{\pi }{2}$.
Again,$0 < \alpha < \frac{\pi }{2}$ and $0 < 2\beta < \frac{\pi }{2}$ both
==> $0 < \alpha + 2\beta < \pi $
Thus, $0 < \alpha + 2\beta < \pi $ and $\tan (\alpha + 2\beta ) = 1$ both
==> $\alpha + 2\beta = \frac{\pi }{4} $
$\Rightarrow 2\beta = \frac{\pi }{4} - \alpha $.
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