${{\sqrt 2 \sin \alpha } \over {\sqrt {1 + \cos 2\alpha } }} = {1 \over 7}$ $\Rightarrow$ ${{\sqrt 2 \sin \alpha } \over {\sqrt {2{{\cos }^2}\alpha } }}$ = ${1 \over 7}$ $\Rightarrow$ ${{\sqrt 2 \sin \alpha } \over {\sqrt 2 \cos \alpha }}$ = ${1 \over 7}$ $\Rightarrow$ tan$\alpha$ = ${1 \over 7}$ Also given $\sqrt {{{1 - \cos 2\beta } \over 2}} = {1 \over {\sqrt {10} }}$ $\Rightarrow$ ${{\sqrt 2 \sin \beta } \over {\sqrt 2 }}$ = ${1 \over {\sqrt {10} }}$ $\Rightarrow$ sin $\beta$ = ${1 \over {\sqrt {10} }}$ $\therefore$ tan $\beta$ = ${1 \over 3}$ $\tan 2\beta = {{2\tan \beta } \over {1 - {{\tan }^2}\beta }}$ = ${{2\left( {{1 \over 3}} \right)} \over {1 - {1 \over 9}}}$ = ${3 \over 4}$ $\therefore$ tan($\alpha$ + 2$\beta$) = ${{\tan \alpha + \tan 2\beta } \over {1 - \tan \alpha .\tan 2\beta }}$ = ${{{1 \over 7} + {3 \over 4}} \over {1 - {1 \over 7}.{3 \over 4}}}$ = ${{25} \over {25}}$ = 1