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

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(A) Given $\frac{\sqrt{2} \sin \alpha}{\sqrt{1+\cos 2 \alpha}}=\frac{1}{7}$. Since $1+\cos 2 \alpha = 2 \cos^2 \alpha$,we have $\frac{\sqrt{2} \sin \a

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(A) Given $\frac{\sqrt{2} \sin \alpha}{\sqrt{1+\cos 2 \alpha}}=\frac{1}{7}$. Since $1+\cos 2 \alpha = 2 \cos^2 \alpha$,we have $\frac{\sqrt{2} \sin \alpha}{\sqrt{2} \cos \alpha} = 	an \alpha = \frac{1}{7}$.Given $\sqrt{\frac{1-\cos 2 \beta}{2}}=\frac{1}{\sqrt{10}}$. Since $1-\cos 2 \beta = 2 \sin^2 \beta$,we have $\sqrt{\sin^2 \beta} = \sin \beta = \frac{1}{\sqrt{10}}$.Since $\sin \beta = \frac{1}{\sqrt{10}}$,then $\cos \beta = \sqrt{1 - \frac{1}{10}} = \frac{3}{\sqrt{10}}$,so $	an \beta = \frac{1}{3}$.Using the formula $	an 2 \beta = \frac{2 	an \beta}{1-	an^2 \beta} = \frac{2(1/3)}{1-(1/9)} = \frac{2/3}{8/9} = \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$.

List-$A$	List-$B$
$(I)$ $\tan \left(\frac{\alpha + \beta}{2}\right) =$	$(a)$ $\frac{b}{a}$
$(II)$ $\cos (\alpha + \beta) =$	$(b)$ $\frac{2ab}{a^2 + b^2}$
$(III)$ $\sin (\alpha + \beta) =$	$(c)$ $\frac{2ab}{a^2 - b^2}$
$(IV)$ $\tan (\alpha + \beta) =$	$(d)$ $\frac{a^2 - b^2}{a^2 + b^2}$

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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