If ${\tan }^{-1}\left(\frac{\mathrm{x}-1}{\mathrm{x}+2}\right)+{\tan }^{-1}\left(\frac{\mathrm{x}+1}{\mathrm{x}+2}\right)=\frac{\mathrm{\pi }}{4}$ then x in is equal to

We have,

${\tan }^{-1}\frac{\mathrm{x}-1}{\mathrm{x}+2}+{\tan }^{-1}\frac{\mathrm{x}+1}{\mathrm{x}+2}=\frac{\mathrm{\pi }}{4}$

$\Rightarrow {\tan }^{-1}\left[\frac{\frac{\mathrm{x}-1}{\mathrm{x}+2}+\frac{\mathrm{x}+1}{\mathrm{x}+2}}{1-\left(\frac{\mathrm{x}-1}{\mathrm{x}+2}\right)\left(\frac{\mathrm{x}+1}{\mathrm{x}+2}\right)}\right]=\frac{\mathrm{\pi }}{4}$

$\begin{array}{l}\Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\left[\frac{2\mathrm{x}(\mathrm{x}+2)}{{\mathrm{x}}^2+4+4\mathrm{x}-{\mathrm{x}}^2+1}\right]\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}=\tan \frac{\mathrm{\pi }}{4}\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\frac{2\mathrm{x}(\mathrm{x}+2)}{4\mathrm{x}+5}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}=1\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}2{\mathrm{x}}^2+4\mathrm{x}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}=4\mathrm{x}+5\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{x}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}=\pm \sqrt{\frac{5}{2}}\end{array}$