Question

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

Answer 1

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

$\sqrt{\frac{1-\cos 2\beta }{2}}=\frac{1}{\sqrt{10}}$
$\Rightarrow \frac{\sqrt{2}\sin \beta }{\sqrt{2}}=\frac{1}{\sqrt{1}0}$
$\Rightarrow \sin \beta =\frac{1}{\sqrt{10}}\Rightarrow \tan \beta =\frac{1}{3}$

$\tan 2\beta =\frac{2\tan \beta }{1-{\tan }^2\beta }=\frac{\frac{2}{3}}{1-\frac{1}{9}}=\frac{3}{4}$

$\tan (\alpha +2\beta )=\frac{\tan \alpha +\tan 2\beta }{1-\tan \alpha \tan 2\beta }\tan (\alpha +2\beta )=\frac{\frac{1}{7}+\frac{3}{4}}{1-\frac{1}{7}\times \frac{3}{4}}=1$