If ${\tan }^{-1}(x+1)+{\cot }^{-1}(x-1)={\sin }^{-1}\frac{4}{5}+{\cos }^{-1}\frac{3}{5}$, then x  has the value : 

$\begin{array}{l}{\tan }^{-1}(x+1)+{\cot }^{-1}(x-1)={\sin }^{-1}\left(\frac{4}{5}\right)+{\cos }^{-1}\left(\frac{3}{5}\right)\end{array}$

$\Rightarrow {\tan }^{-1}(x+1)+{\tan }^{-1}\left(\frac{1}{x-1}\right)={\tan }^{-1}\frac{4}{3}+{\tan }^{-1}\frac{4}{3}=2{\tan }^{-1}\frac{4}{3}$

$\begin{array}{l}\Rightarrow {\tan }^{-1}\left[\frac{x+1+\frac{1}{x-1}}{1-(x+1)\left(\frac{1}{x-1}\right)}\right]={\tan }^{-1}\left[\frac{2\times \frac{4}{3}}{1-\frac{16}{9}}\right] \left[\text{ using }2{\tan }^{-1}x={\tan }^{-1}\left(\frac{2x}{1-x^2}\right)\right.\end{array}$

$\begin{array}{l}\Rightarrow {\tan }^{-1}\left(-\frac{x^2}{2}\right)={\tan }^{-1}\left(-\frac{24}{7}\right)\\ \Rightarrow \frac{x^2}{2}=\frac{24}{7}\Rightarrow x^2=\frac{48}{7}\Rightarrow x=4\sqrt{\frac{3}{7}}\end{array}$