Question

If $\tan \left(\alpha \right)=\frac{1}{7},\tan \left(\beta \right)=\frac{1}{3}$, then $\cos \left(2\alpha \right)=$

• A. $\sin \left(2\beta \right)$
• B. $\sin \left(4\beta \right)$
• C. $\sin \left(3\beta \right)$
• D. None of these

Answer 1

The correct option is B

$\sin \left(4\beta \right)$

Explanation for the correct option:

$\tan \left(\alpha \right)=\frac{1}{7},\tan \left(\beta \right)=\frac{1}{3}$

$\begin{array}{rcl}\tan \left(2\beta \right)& =& \frac{2\tan \left(\beta \right)}{1-{\tan }^2\left(\beta \right)}\\ & =& \frac{2\left({\displaystyle \frac{1}{3}}\right)}{1-{\left({\displaystyle \frac{1}{3}}\right)}^2}\\ & =& \frac{{\displaystyle \frac{2}{3}}}{1-{\displaystyle \frac{1}{9}}}\\ & =& \frac{{\displaystyle \frac{2}{3}}}{{\displaystyle \frac{8}{9}}}\\ & =& \frac{3}{4}\end{array}$

Now $\tan \left(\alpha +2\beta \right)=\frac{\tan \left(\alpha \right)+\tan \left(2\beta \right)}{1-\tan \left(\alpha \right)\tan \left(2\beta \right)}$

Then,

$\begin{array}{rcl}\tan \left(\alpha +2\beta \right)& =& \frac{{\displaystyle \frac{1}{7}}+{\displaystyle \frac{3}{4}}}{1-\left({\displaystyle \frac{1}{7}}\right)\left({\displaystyle \frac{3}{4}}\right)}\\ & =& \frac{{\displaystyle \frac{25}{28}}}{{\displaystyle \frac{25}{28}}}\\ & =& 1\end{array}$

$\begin{array}{rcl}\therefore \tan \left(\alpha +2\beta \right)& =& 1\\ \tan \left(\alpha +2\beta \right)& =& \tan \left(\frac{\pi }{4}\right)\\ \alpha +2\beta & =& \frac{\pi }{4}\\ \alpha & =& \frac{\pi }{4}-2\beta \\ 2\alpha & =& \frac{\pi }{2}-4\beta \\ \cos \left(2\alpha \right)& =& \cos \left(\frac{\pi }{2}-4\beta \right)\\ \therefore \cos \left(2\alpha \right)& =& \sin \left(4\beta \right)\end{array}$

Hence, the correct option is (B).