If tanα=

It is given that $\tan \alpha=\frac{1}{7}$ and $\tan \beta=\frac{1}{3}$.

Now,

$\tan 2 \beta=\frac{2 \tan \beta}{1-\tan ^{2} \beta}$

$=\frac{2 \times \frac{1}{3}}{1-\frac{1}{9}}$

$=\frac{\frac{2}{3}}{\frac{8}{9}}$

$=\frac{3}{4}$

$\therefore \tan (\alpha+2 \beta)=\frac{\tan \alpha+\tan 2 \beta}{1-\tan \alpha \tan 2 \beta}$

$=\frac{\frac{1}{7}+\frac{3}{4}}{1-\frac{1}{7} \times \frac{3}{4}}$

$=\frac{\frac{25}{28}}{\frac{25}{28}}$

$=1$

$\tan (\alpha+2 \beta)=1=\tan \frac{\pi}{4}$

$\Rightarrow \alpha+2 \beta=\frac{\pi}{4}$

$\Rightarrow \alpha=\frac{\pi}{4}-2 \beta$

$\Rightarrow 2 \alpha=\frac{\pi}{2}-4 \beta$

$\Rightarrow \cos 2 \alpha=\cos \left(\frac{\pi}{2}-4 \beta\right)=\sin 4 \beta$

$\therefore \cos 2 \alpha=\sin 4 \beta$

Hence, the correct answer is option $B$.