चूँकि $\sin \beta  = \frac{1}{{\sqrt {10} }} \Rightarrow \tan \beta  = \frac{1}{3}$

$\Rightarrow$ $\tan 2\beta  = \frac{{2\tan \beta }}{{1 – {{\tan }^2}\beta }} = \frac{3}{4}$

$\therefore \tan (\alpha  + 2\beta ) = \frac{{\frac{1}{7} + \frac{3}{4}}}{{1 – \frac{1}{7}.\frac{3}{4}}} = \frac{{25}}{{25}} = 1$

अब $0 < \beta  < \frac{\pi }{2}$ एवं $\tan 2\beta  = \frac{3}{4} > 0$ दोनों

$\Rightarrow$  $0 < 2\beta  < \frac{\pi }{2}$.

पुन: $0 < \alpha  < \frac{\pi }{2}$ एवं $0 < 2\beta  < \frac{\pi }{2}$ दोनों 

$\Rightarrow$  $0 < \alpha  + 2\beta  < \pi $

अत: $0 < \alpha  + 2\beta  < \pi $ एवं $\tan (\alpha  + 2\beta ) = 1$ दोनों     

$\Rightarrow$  $\alpha  + 2\beta  = \frac{\pi }{4} $

$\Rightarrow 2\beta  = \frac{\pi }{4} – \alpha $.