Question

The value of $\sin \left(4{\tan }^{-1}\frac{1}{3}\right)-\cos \left(2{\tan }^{-1}\frac{1}{7}\right)$ is

• A. $\frac{4}{7}$
• B. $0$
• C. $\frac{7}{8}$
• D. None of these

Answer 1

The correct option is B $0$

$\sin \left(4{\tan }^{-1}\frac{1}{3}\right)=2\sin \left(2{\tan }^{-1}\frac{1}{3}\right)\cos \left(2{\tan }^{-1}\frac{1}{3}\right)$
$=2\sin \left({\tan }^{-1}\frac{3}{4}\right)\cos \left({\tan }^{-1}\frac{3}{4}\right)$
$[\because 2{\tan }^{-1}\frac{1}{3}={\tan }^{-1}\left(\frac{2.\frac{1}{3}}{1-{\left(\frac{1}{3}\right)}^2}\right)={\tan }^{-1}\frac{3}{4}]$
$=2.\frac{3}{5}.\frac{4}{5}=\frac{24}{25}$
Also, $\cos \left(2{\tan }^{-1}\frac{1}{7}\right)=\cos \left({\tan }^{-1}\frac{7}{24}\right)=\frac{24}{25}$
$\therefore$ The given expression $=0$