Prove the following

Question:

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

$\alpha, \beta \in\left(0, \frac{\pi}{2}\right)$, then $\tan (\alpha+2 \beta)$ is equal to

Solution:

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

$\Rightarrow \quad \frac{\sqrt{2} \sin \beta}{\sqrt{2}}=\frac{1}{\sqrt{10}}$

$\therefore \quad \tan \alpha=\frac{1}{7}$ and $\sin \beta=\frac{1}{\sqrt{10}}$

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

$\therefore \quad \tan 2 \beta=\frac{2 \tan \beta}{1-\tan ^{2} \beta}=\frac{2 \cdot \frac{1}{3}}{1-\frac{1}{9}}=\frac{\frac{2}{3}}{\frac{8}{9}}=\frac{3}{4}$

$\tan (\alpha+2 \beta)=\frac{\tan \alpha+\tan 2 \beta}{1-\tan \alpha \tan 2 \beta}$

$=\frac{\frac{1}{7}+\frac{3}{4}}{1-\frac{1}{7} \cdot \frac{3}{4}}=\frac{\frac{4+21}{28}}{\frac{25}{28}}=1$

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