Mathematics - Inverse Trigonometric Functions Question with Solution | TestHub

$\begin{array}{rl}& 2{\tan }^{-1}\frac{1}{5}={\tan }^{-1}\frac{1}{5}+{\tan }^{-1}\frac{1}{5}\\ & ={\tan }^{-1}\frac{\frac{1}{5}+\frac{1}{5}}{1-\frac{1}{5}\cdot \frac{1}{5}}={\tan }^{-1}\frac{2/5}{24/25}={\tan }^{-1}\frac{5}{12}\\ & =\tan \left(2{\tan }^{-1}\frac{1}{5}-\frac{\pi }{4}\right)=\tan \left({\tan }^{-1}\frac{5}{12}-\frac{\pi }{4}\right)\\ & =\frac{\tan \left({\tan }^{-1}\frac{5}{12}\right)-\tan \frac{\pi }{4}}{1+\tan \left({\tan }^{-1}\frac{5}{12}\right)\tan \frac{\pi }{4}}=\frac{-\frac{7}{12}}{\frac{17}{12}}=-\frac{7}{17}\end{array}$