The value of $6\tan \left({\cos }^{-1}\frac{4}{5}+{\tan }^{-1}\frac{2}{3}\right),$ is

We have,

    $\begin{array}{r}{\cos }^{-1}\left(\frac{4}{5}\right)={\tan }^{-1}\left(\frac{3}{4}\right)\\ \therefore 6\tan \left({\cos }^{-1}\frac{4}{5}+{\tan }^{-1}\frac{2}{3}\right)\end{array}$

             $\begin{array}{l}=6\tan \left({\tan }^{-1}\frac{3}{4}+{\tan }^{-1}\frac{2}{3}\right)\\ =6\tan \left\{{\tan }^{-1}\left(\frac{\frac{3}{4}+\frac{2}{3}}{1-\frac{3}{4}\times \frac{2}{3}}\right)\right\}\\ =6\tan \left\{{\tan }^{-1}\left(\frac{17}{6}\right)\right\}=17\end{array}$