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

The correct option is B

$sin4\beta$

Given that:

$tan\alpha =\frac{1}{7},tan\beta =\frac{1}{3}cos2\alpha =\frac{1-\frac{1}{49}}{1+\frac{1}{49}}=\frac{\frac{48}{49}}{\frac{50}{49}}=\frac{48}{50}=\frac{24}{25}$
$\Rightarrow cos2\alpha =\frac{24}{25}$ . . .(i)

We know that, $sin4\beta =\frac{2tan2\beta }{1+ta{n}^{2}2\beta }$ . . . (ii)

and $tan2\beta =\frac{2tan\beta }{1-ta{n}^2\beta }=\frac{2\times \frac{1}{3}}{1-\frac{1}{9}}$

$=\frac{\frac{2}{3}}{\frac{8}{9}}=\frac{2\times 9}{3\times 8}=\frac{3}{4}$