If $\tan \alpha =1/7\text{ and }\sin \beta =1/\sqrt{10}$ where $0<\alpha ,\beta <\pi /2$ then $2\beta$   is equal to

$\sin \beta =\frac{1}{\sqrt{10}}\Rightarrow \cos \beta =\sqrt{1-\frac{1}{10}}=\frac{3}{\sqrt{10}}$

$\begin{array}{l}\Rightarrow \tan \beta =1/3\\ \therefore \tan 2\beta =\frac{2\tan \beta }{1-{\tan }^2\beta }=\frac{2\cdot 1/3}{1-1/9}=\frac{3}{4}\end{array}$

and  $\tan (\alpha +2\beta )$

  $-\frac{\tan \alpha +\tan 2\beta }{1-\tan \alpha \tan 2\beta }-\frac{1/7+3/4}{1-(1/7)(3/4)}-\frac{25}{25}-1$

Since  $0<\beta <\pi /2$  and  $\tan 2\beta =3/4>0$  wet get

$0<2\beta <\pi /2.$$0<\alpha <\pi /4$   Hence, $0<\alpha +2\beta <\pi$

and $\tan (\alpha +2\beta )=1$   so that 

 $\alpha +2\beta =\pi /4\Rightarrow 2\beta =\pi /4-\alpha$