If ${\tan }^{-1}(\mathrm{x}+1)+{\cot }^{-1}\left(\frac{1}{\mathrm{x}-1}\right)={\tan }^{-1}\left(\frac{8}{31}\right)$then which are the following is/are TRUE?

${\tan }^{-1}(\mathrm{x}+1)+{\cot }^{-1}\left(\frac{1}{\mathrm{x}-1}\right)={\tan }^{-1}\left(\frac{8}{31}\right)$
${\tan }^{-1}(\mathrm{x}+1)+ta{n}^{-1}\left(x-1\right)={\tan }^{-1}\left(\frac{8}{31}\right)$
$$\begin{gathered} \Rightarrow {\tan }^{-1}\frac{\mathrm{x}+1+x-1}{1-x^2+1}={\tan }^{-1}\left(\frac{8}{31}\right) \\ \Rightarrow \frac{2x}{2-x^2}=\frac{8}{31} \\ \Rightarrow 31x=8-4x^2\Rightarrow 4x^2-31x-8=0 \\ \Rightarrow (4x-1)(x+8)=0 \end{gathered}$$

Solve we set $\mathrm{x}=\frac{1}{4},-8\text{, if }\mathrm{x}=\frac{1}{4},\text{ LHS }>\frac{\mathrm{\pi }}{2},\text{ RHS }<\frac{\mathrm{\pi }}{2}$