Using double angle identity

\cos 2β = 1-2 \sin^2β

= 1-2\Big( \dfrac{1}{\sqrt{10} } \Big) ^2

=\dfrac{4}{5}

Then

\sin 2β =\dfrac{3}{5}

\begin{aligned} \tanα=\dfrac{1}{7} \implies &\sinα = \dfrac{1}{\sqrt{1+7^2} } = \dfrac{1}{\sqrt{50} } = \dfrac{1}{5\sqrt{2} } \\ &\cos α = \dfrac{7}{5\sqrt{2} } \end{aligned}

\cos( α +2β ) = \cos α\cos 2β -\sin α \sin 2β

= \dfrac{7}{5\sqrt{2} }\cdot \dfrac{4}{5} - \dfrac{1}{5\sqrt{2} }\cdot \dfrac{3}{5}

=\dfrac{\sqrt{2} }{2}

Therefore

α +2β = 45\degree

C is the correct choice