Question

If $tan\alpha=\frac{1}{7}$, $tan\beta=\frac{1}{3}$, then $cos2\alpha$ is equal to

• A. $sin2\beta$
• B. $sin4\beta$
• C. $sin3\beta$
• D. $cos2\beta$

Answer 1

Answer: (B)

We have, $tan\alpha=\frac{1}{7}$,
$tan\beta=\frac{1}{3}$
Now, $cos2\alpha=\frac{1-tan^{2}\,\alpha}{1+tan^{2}\,\alpha}$
$=\frac{1-\frac{1}{49}}{1+\frac{1}{49}}=\frac{48}{50}\quad\ldots\left(i\right)$

Now, $sin2\beta=\frac{2\,tan\,\beta}{1+tan^{2}\,\beta}$
$=\frac{2\times \frac{1}{3}}{1+\frac{1}{9}}=\frac{2}{3}\times\frac{9}{10}=\frac{3}{5}\quad\ldots\left(ii\right)$

Also, $sin4\beta=2sin2\beta\,cos2\beta=2\cdot \frac{3}{5}\left(\frac{1-tan^{2}\,\beta}{1+tan^{2}\,\beta}\right)$
(using $(ii)$)
$=\frac{6}{5}\left(\frac{1-\frac{1}{9}}{1+\frac{1}{9}}\right)=\frac{6}{5}\times\frac{8}{9}\times\frac{9}{10}=\frac{48}{50}\quad\ldots\left(iii\right)$
Hence, $cos\, 2\alpha = sin\, 4\beta$ (from $(i)$ and $(iii)$)