If $\tan \alpha=\frac{1}{7}, \tan \beta=\frac{1}{3}$, then $\cos 2 \alpha$ is equal to
(a) $\sin 2 \beta$
(b) $\sin 4 \beta$
(c) $\sin 3 \beta$
(d) $\cos 2 \beta$
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 $B$.
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