MCQ
If $\tan\alpha=\frac{1}{7},\tan\beta=\frac{1}{3},$ then $\cos2\alpha$ is equal to:
  • A
    $\sin2\beta$
  • $\sin4\beta$
  • C
    $\sin3\beta$
  • D
    $\cos2\beta$

Answer

Correct option: B.
$\sin4\beta$
Given that, $\tan\alpha=\frac{1}{7}$ and $\tan\beta=\frac{1}{3}$
$\cos2\alpha=\frac{1-\tan^2\alpha}{1+\tan^2\alpha}=\frac{1-\Big(\frac{1}{7}\Big)^2}{1+\Big(\frac{1}{7}\Big)^2}=\frac{1-\frac{1}{49}}{1+\frac{1}{49}}$
$=\frac{48}{50}=\frac{24}{25}$
Now $\tan2\beta=\frac{2\tan\beta}{1-\tan^2\beta}=\frac{2\times\frac{1}{3}}{1-\frac{1}{9}}=\frac{\frac{2}{3}}{\frac{8}{9}}=\frac{2}{3}\times\frac{9}{8}=\frac{3}{4}$
$\therefore\tan^2\beta=\frac{3}{4}$
$\sin4\beta=\frac{2\tan2\beta}{1+\tan^22\beta}$
$=\frac{2\times\frac{3}{4}}{1+\Big(\frac{3}{4}\Big)^2}$
$=\frac{\frac{3}{2}}{1+\frac{9}{16}}$
$=\frac{3}{2}\times\frac{16}{25}$
$=\frac{24}{25}$
$\cos2\alpha=\sin4\beta=\frac{24}{25}$
Hence, the correct option is $(b).$

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