If 2 sinα 1 + 2α = (1)/(7) and (1 - 2β)/(2) = 1 10 , α, β ∈ (0,...

Given,

$\dfrac{\sqrt{2}sin\alpha}{\sqrt{1 + \cos 2\alpha}} = \dfrac{1}{7}$ and $\sqrt{\dfrac{1 - \cos 2\beta}{2}} = \dfrac{1}{\sqrt{10}}, \alpha, \beta \in \left(0, \dfrac{\pi}{2}\right)$

Now using $\cos 2\theta = 2\cos^{2}\theta - 1 = 1 - 2\sin^{2}\theta$, we can write

$\dfrac{\sqrt{2}sin \alpha}{\sqrt{2}cos \alpha} = \dfrac{1}{7}$ and $\dfrac{\sqrt{2}sin \beta}{\sqrt{2}} = \dfrac{1}{\sqrt{10}}$

$tan \alpha = \dfrac{1}{7}$ and $sin \beta = \dfrac{1}{\sqrt{10}}$ or $tan \beta = \dfrac{1}{3}$
$\tan 2\beta = \dfrac{2tan \beta}{1 - \tan^{2} \beta} = 2 . \dfrac{\left(\dfrac{1}{3}\right)}{\left[1 - \left(\dfrac{1}{9}\right)\right]} = \dfrac{3}{4}$
$\tan \left(\alpha + 2\beta\right) = \dfrac{tan\alpha + \tan 2\beta}{1 - tan\alpha tan 2\beta} = \dfrac{\dfrac{1}{7} + \dfrac{3}{4}}{1 - \dfrac{1}{7} \cdot \dfrac{3}{4}} = \dfrac{\dfrac{4 + 21}{28}}{\dfrac{25}{28}} = 1$