The value of $\tan \left(2{\tan }^{-1}\left(\frac{3}{5}\right)+{\sin }^{-1}\left(\frac{5}{13}\right)\right)$ is equal to

Let $\mathrm{A}=\tan \left(2{\tan }^{-1}\left(\frac{3}{5}\right)+{\sin }^{-1}\left(\frac{5}{13}\right)\right)$            ...(i)

Now, using $2{\tan }^{-1}\mathrm{x}={\tan }^{-1}\frac{2\mathrm{x}}{1-{\mathrm{x}}^2}$

$\begin{array}{c}2{\tan }^{-1}\left(\frac{3}{5}\right)={\tan }^{-1}\left(\frac{2\left(\frac{3}{5}\right)}{1-{\left(\frac{3}{5}\right)}^2}\right)={\tan }^{-1}\left(\frac{\frac{6}{5}}{1-\frac{9}{25}}\right)\\ ={\tan }^{-1}\left(\frac{30}{16}\right)={\tan }^{-1}\left(\frac{15}{8}\right)\end{array}$

Let ${\sin }^{-1}\left(\frac{5}{13}\right)=\mathrm{\theta },\text{ then }\sin \mathrm{\theta }=\frac{5}{13}$

$\begin{array}{l}\therefore \tan \mathrm{\theta }=\frac{\sin \mathrm{\theta }}{\sqrt{1-{\sin }^2\mathrm{\theta }}}=\frac{\frac{5}{13}}{\sqrt{1-{\left(\frac{5}{3}\right)}^2}}=\frac{\frac{5}{13}}{\frac{12}{13}}\\ \Rightarrow \mathrm{\theta }={\tan }^{-1}\left(\frac{5}{12}\right)={\sin }^{-1}\left(\frac{5}{13}\right)\end{array}$

From Eq. (i),

$\begin{array}{c}\mathrm{A}=\tan \left({\tan }^{-1}\left(\frac{15}{8}\right)+{\tan }^{-1}\left(\frac{5}{12}\right)\right)\\ =\tan \left[{\tan }^{-1}\left(\frac{\frac{15}{8}+\frac{5}{12}}{1-\left(\frac{15}{8}\right)\left(\frac{5}{12}\right)}\right]\right]\\ \begin{array}{c}=\tan \left[{\tan }^{-1}\left(\frac{\frac{55}{24}}{\frac{21}{96}}\right)\right]=\tan \left[{\tan }^{-1}\frac{55\times 4}{21}\right]\\ =\tan \left({\tan }^{-1}\frac{220}{21}\right)=\frac{220}{21}\end{array}\end{array}$