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

$\frac{\sqrt{2}\sin \alpha }{\sqrt{1+\cos 2\alpha }}=\frac{1}{7}\hspace{0.17em}$

$\frac{\sqrt{2}\sin \alpha }{\sqrt{2{\cos }^2\alpha }}=\frac{1}{7}$

$\frac{\sqrt{2}\sin \alpha }{\sqrt{2}\cos \alpha }=\frac{1}{7}\Rightarrow \tan \alpha =\frac{1}{7}$

$\frac{\sqrt{1-\cos 2\beta }}{2}=\frac{1}{\sqrt{10}}$

$\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\frac{\sqrt{2{\sin }^2\beta }}{2}=\frac{1}{\sqrt{10}}$

$\sin \beta =\frac{1}{\sqrt{10}}$

$\tan \beta =\frac{1}{3}$

$\tan \left(\alpha +\beta \right)=\frac{\tan \alpha +\tan \beta }{1-\tan \alpha \tan \beta }$

$\begin{array}{l}\tan \left(\alpha +\beta \right)==\frac{\frac{1}{7}+\frac{1}{3}}{1-\left(\frac{1}{7}\right)\left(\frac{1}{3}\right)}\\ =\frac{3+7}{21-1}=\frac{10}{20}=\frac{1}{2}\end{array}$