If tan alpha = frac{1}{7} and tan beta = frac | vedclass

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(B) Given $	an \alpha = \frac{1}{7}$.Using the formula $\cos 2\alpha = \frac{1 - 	an^2 \alpha}{1 + 	an^2 \alpha}$:$\cos 2\alpha = \frac{1 - (1/7)^2

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$\sin 4\beta$

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(B) Given $	an \alpha = \frac{1}{7}$.Using the formula $\cos 2\alpha = \frac{1 - 	an^2 \alpha}{1 + 	an^2 \alpha}$:$\cos 2\alpha = \frac{1 - (1/7)^2}{1 + (1/7)^2} = \frac{1 - 1/49}{1 + 1/49} = \frac{48/49}{50/49} = \frac{48}{50} = \frac{24}{25}$.Given $	an \beta = \frac{1}{3}$.Using the formula $\sin 2\beta = \frac{2 	an \beta}{1 + 	an^2 \beta}$:$\sin 2\beta = \frac{2(1/3)}{1 + (1/3)^2} = \frac{2/3}{1 + 1/9} = \frac{2/3}{10/9} = \frac{2}{3} 	imes \frac{9}{10} = \frac{3}{5}$.Since $\sin 2\beta = \frac{3}{5}$,then $\cos 2\beta = \sqrt{1 - (3/5)^2} = \frac{4}{5}$.Now,$\sin 4\beta = 2 \sin 2\beta \cos 2\beta = 2 	imes \frac{3}{5} 	imes \frac{4}{5} = \frac{24}{25}$.Thus,$\cos 2\alpha = \sin 4\beta$.

If $\tan \alpha = \frac{1}{7}$ and $\tan \beta = \frac{1}{3}$,then $\cos 2\alpha = $

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