Question

If $\tan \left(A\right)=\frac{1}{2},\tan \left(B\right)=\frac{1}{3}$, then $\cos \left(2A\right)=$

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

Answer 1

The correct option is B

$\sin \left(2B\right)$

Solving the given function:

$\tan \left(A\right)=\frac{1}{2},\tan \left(B\right)=\frac{1}{3}$

We know that,

$\cos \left(2A\right)=\frac{1-{\tan }^2\left(A\right)}{1+{\tan }^2\left(A\right)}$

Substitute value of $\tan \left(A\right)$

$\begin{array}{rcl}\cos \left(2A\right)& =& \frac{1-{\left({\displaystyle \frac{1}{2}}\right)}^2}{1+{\left({\displaystyle \frac{1}{2}}\right)}^2}\\ & =& \frac{1-{\displaystyle \frac{1}{4}}}{1+{\displaystyle \frac{1}{4}}}\\ & =& \frac{4-1}{4+1}\\ & =& \frac{3}{5}\end{array}$

$\sin \left(2B\right)=\frac{2\tan \left(B\right)}{1+{\tan }^2\left(B\right)}$

Substitute values of $\tan \left(B\right)$

$\begin{array}{rcl}\sin \left(2B\right)& =& \frac{2\left({\displaystyle \frac{1}{3}}\right)}{1+{\left({\displaystyle \frac{1}{3}}\right)}^2}\\ & =& \frac{{\displaystyle \frac{2}{3}}}{1+{\displaystyle \frac{1}{9}}}\\ & =& \frac{{\displaystyle \frac{2}{3}}}{{\displaystyle \frac{10}{9}}}\\ & =& \frac{3}{5}\end{array}$

Therefore, $\cos \left(2A\right)=\sin \left(2B\right)$

Hence, the correct option is (B).