Mathematics - Inverse Trigonometric Function Question with Solution | TestHub

$\sin \left(4{\tan }^{-1}\frac{1}{3}\right)$ $\begin{array}{rl}& =2\sin \left(2{\tan }^{-1}\frac{1}{3}\right)\cos \left(2{\tan }^{-1}\frac{1}{3}\right)\\ & =2\sin \left({\tan }^{-1}\frac{3}{4}\right)=\cos \left({\tan }^{-1}\frac{3}{4}\right)=2\cdot \frac{3}{5}\cdot \frac{4}{5}=\frac{24}{25}\\ & \cos \left(2{\tan }^{-1}\frac{1}{7}\right)=\cos \left({\tan }^{-1}\frac{7}{24}\right)=\frac{24}{25}.\end{array}$ The given expression $=0$