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If two angles $\alpha, \beta$ are such that $0 < \alpha, \beta < \frac{\pi}{4}$, $\sqrt{1+\cos 2 \alpha}=\frac{3}{\sqrt{5}}$ and $\frac{\sqrt{1-\cos 2 \beta}}{1+\cos 2 \beta}=\frac{1}{7}$, then $(2 \alpha+\beta)=$
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The correct answer is:
$\frac{\pi}{4}$
Given,
$\sqrt{1+\cos 2 \alpha}=\frac{3}{\sqrt{5}}$ and $\sqrt{\frac{1-\cos 2 \beta}{1+\cos 2 \beta}}=\frac{1}{7}$
$\begin{aligned} & \Rightarrow \quad \sqrt{2} \cos \alpha=\frac{3}{\sqrt{5}} \text { and } \frac{\sin \beta}{\cos \beta}=\frac{1}{7} \\ & \Rightarrow \quad \cos \alpha=\frac{3}{\sqrt{10}} \text { and } \tan \beta=\frac{1}{7} \\ & \Rightarrow \cos 2 \alpha=2 \times \frac{9}{10}-1\end{aligned}$
$\Rightarrow \cos 2 \alpha=\frac{8}{10}=\frac{4}{5} \Rightarrow \tan 2 \alpha=\frac{\frac{3}{5}}{\frac{4}{5}}=\frac{3}{4}$
$\tan (2 \alpha+\beta)=\frac{\frac{3}{4}+\frac{1}{7}}{1-\frac{3}{28}}=\frac{25}{25}=1 \Rightarrow 2 \alpha+\beta=\frac{\pi}{4}$
$\sqrt{1+\cos 2 \alpha}=\frac{3}{\sqrt{5}}$ and $\sqrt{\frac{1-\cos 2 \beta}{1+\cos 2 \beta}}=\frac{1}{7}$
$\begin{aligned} & \Rightarrow \quad \sqrt{2} \cos \alpha=\frac{3}{\sqrt{5}} \text { and } \frac{\sin \beta}{\cos \beta}=\frac{1}{7} \\ & \Rightarrow \quad \cos \alpha=\frac{3}{\sqrt{10}} \text { and } \tan \beta=\frac{1}{7} \\ & \Rightarrow \cos 2 \alpha=2 \times \frac{9}{10}-1\end{aligned}$
$\Rightarrow \cos 2 \alpha=\frac{8}{10}=\frac{4}{5} \Rightarrow \tan 2 \alpha=\frac{\frac{3}{5}}{\frac{4}{5}}=\frac{3}{4}$
$\tan (2 \alpha+\beta)=\frac{\frac{3}{4}+\frac{1}{7}}{1-\frac{3}{28}}=\frac{25}{25}=1 \Rightarrow 2 \alpha+\beta=\frac{\pi}{4}$
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