MCQ
If $\tan \alpha=\frac{1}{7}$ and $\tan \beta=\frac{1}{3}$, then $\cos 2 \alpha=$
  • A
    $\sin 2 \beta$
  • $\sin 4 \beta$
  • C
    $\sin 3 \beta$
  • D
    $\sin \beta$

Answer

Correct option: B.
$\sin 4 \beta$
(B)
$\cos 2 \alpha=\frac{1-t^2}{1+t^2}=\frac{24}{25}($ Let $t=\tan \alpha)$
$\sin 2 \beta-\frac{2 T}{1+T^2}-\frac{3}{5}(\operatorname{Let} T-\tan \beta)$
$\Rightarrow \cos 2 \beta=\frac{4}{5}$
Now, $\sin 4 \beta=2 \sin 2 \beta \cos 2 \beta$
$=2 \cdot \frac{3}{5}. \frac{4}{5}$
$=\frac{24}{25}$
$=\cos 2 \alpha$

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