If $\frac{\sqrt{2}\mathrm{s}\mathrm{i}\mathrm{n}\alpha }{\sqrt{1+\mathrm{c}\mathrm{o}\mathrm{s}2\alpha }}=\frac{1}{7}$ and $\sqrt{\frac{1-\mathrm{c}\mathrm{o}\mathrm{s}2\beta }{2}}=\frac{1}{\sqrt{10}}, \alpha ,\beta \in \left(0,\frac{\pi }{2}\right)$, then $\mathrm{t}\mathrm{a}\mathrm{n}\left(\alpha +2\beta \right)$, is equal to

Given,

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

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

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

$$
$\tan \alpha =\frac{1}{7}$ and $\sin \beta =\frac{1}{\sqrt{10}}$ or $\tan \beta =\frac{1}{3}$
$tan2\beta =\frac{2\tan \beta }{1-{\tan }^2\beta }=2.\frac{\left({\displaystyle \frac{1}{3}}\right)}{\left[1-\left({\displaystyle \frac{1}{9}}\right)\right]}=\frac{3}{4}$
$\tan \left(\alpha +2\beta \right)=\frac{tan\alpha +tan2\beta }{1-tan\alpha \tan 2\beta }=\frac{\frac{1}{7}+\frac{3}{4}}{1-\frac{1}{7}·\frac{3}{4}}=\frac{\frac{4+21}{28}}{\frac{25}{28}}=1$