If two angles alpha, beta are such that 0

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(D) Given,$\sqrt{1+\cos 2 \alpha}=\frac{3}{\sqrt{5}}$ and $\sqrt{\frac{1-\cos 2 \beta}{1+\cos 2 \beta}}=\frac{1}{7}$.Since $1+\cos 2 \alpha = 2 \cos^2

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$\frac{\pi}{4}$

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(D) Given,$\sqrt{1+\cos 2 \alpha}=\frac{3}{\sqrt{5}}$ and $\sqrt{\frac{1-\cos 2 \beta}{1+\cos 2 \beta}}=\frac{1}{7}$.Since $1+\cos 2 \alpha = 2 \cos^2 \alpha$,we have $\sqrt{2} \cos \alpha = \frac{3}{\sqrt{5}} \Rightarrow \cos \alpha = \frac{3}{\sqrt{10}}$.Then $\cos 2 \alpha = 2 \cos^2 \alpha - 1 = 2(\frac{9}{10}) - 1 = \frac{18}{10} - 1 = \frac{8}{10} = \frac{4}{5}$.Since $\sin^2 2 \alpha = 1 - \cos^2 2 \alpha = 1 - \frac{16}{25} = \frac{9}{25}$,we have $\sin 2 \alpha = \frac{3}{5}$.Thus,$	an 2 \alpha = \frac{\sin 2 \alpha}{\cos 2 \alpha} = \frac{3/5}{4/5} = \frac{3}{4}$.For $\beta$,$\sqrt{\frac{1-\cos 2 \beta}{1+\cos 2 \beta}} = \sqrt{\frac{2 \sin^2 \beta}{2 \cos^2 \beta}} = 	an \beta = \frac{1}{7}$.Now,$	an(2 \alpha + \beta) = \frac{	an 2 \alpha + 	an \beta}{1 - 	an 2 \alpha 	an \beta} = \frac{3/4 + 1/7}{1 - (3/4)(1/7)} = \frac{(21+4)/28}{(28-3)/28} = \frac{25}{25} = 1$.Therefore,$2 \alpha + \beta = 	an^{-1}(1) = \frac{\pi}{4}$.

If two angles $\alpha, \beta$ are such that $0 < \alpha, \beta < \frac{\pi}{4}$,$\sqrt{1+\cos 2 \alpha}=\frac{3}{\sqrt{5}}$ and $\frac{\sqrt{1-\cos 2 \beta}}{\sqrt{1+\cos 2 \beta}}=\frac{1}{7}$,then $(2 \alpha+\beta)=$

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