If frac{sqrt{2} sin alpha}{sqrt{1+cos 2 alpha | vedclass

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(A) Given $\frac{\sqrt{2} \sin \alpha}{\sqrt{1+\cos 2 \alpha}} = \frac{1}{7}$.Using the identity $1+\cos 2 \alpha = 2 \cos^2 \alpha$,we get $\frac{\sq

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(A) Given $\frac{\sqrt{2} \sin \alpha}{\sqrt{1+\cos 2 \alpha}} = \frac{1}{7}$.Using the identity $1+\cos 2 \alpha = 2 \cos^2 \alpha$,we get $\frac{\sqrt{2} \sin \alpha}{\sqrt{2 \cos^2 \alpha}} = \frac{\sqrt{2} \sin \alpha}{\sqrt{2} \cos \alpha} = 	an \alpha = \frac{1}{7}$.Given $\sqrt{\frac{1-\cos 2 \beta}{2}} = \frac{1}{\sqrt{10}}$.Using the identity $1-\cos 2 \beta = 2 \sin^2 \beta$,we get $\sqrt{\frac{2 \sin^2 \beta}{2}} = \sin \beta = \frac{1}{\sqrt{10}}$.Since $\sin \beta = \frac{1}{\sqrt{10}}$,we have $\cos \beta = \sqrt{1 - \frac{1}{10}} = \frac{3}{\sqrt{10}}$,so $	an \beta = \frac{1}{3}$.Now,$	an 2 \beta = \frac{2 	an \beta}{1 - 	an^2 \beta} = \frac{2(1/3)}{1 - (1/9)} = \frac{2/3}{8/9} = \frac{2}{3} 	imes \frac{9}{8} = \frac{3}{4}$.Finally,$	an (\alpha + 2 \beta) = \frac{	an \alpha + 	an 2 \beta}{1 - 	an \alpha 	an 2 \beta} = \frac{1/7 + 3/4}{1 - (1/7)(3/4)} = \frac{(4+21)/28}{(28-3)/28} = \frac{25}{25} = 1$.

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

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