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, $cos2\alpha =\frac{1-ta{n}^2\alpha }{1+ta{n}^2\alpha }$
$=\frac{1-\frac{1}{49}}{1+\frac{1}{49}}=\frac{48}{50}\dots \left(i\right)$

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

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