Considering only the principal values of inverse trigonometric functions, the number of positive real values of $x$ satisfying ${\tan }^{-1}\left(\mathrm{x}\right)+{\tan }^{-1}\left(2\mathrm{x}\right)=\frac{\mathrm{\pi }}{4}$ is :

Let, 

$\mathrm{f}\left(\mathrm{x}\right)={\tan }^{-1}\mathrm{x}+{\tan }^{-1}2\mathrm{x}$

Now, differentiating the above function we get,

$\Rightarrow {\mathrm{f}}^{\text{'}}\left(\mathrm{x}\right)=\frac{1}{1+{\mathrm{x}}^2}+\frac{2}{1+4{\mathrm{x}}^2}$

$\Rightarrow {\mathrm{f}}^{\text{'}}\left(\mathrm{x}\right)>0; \forall \mathrm{x}\in \mathrm{R}$

$\Rightarrow \underset{\mathrm{x}\to \infty }{\lim }\mathrm{f}\left(\mathrm{x}\right)=\mathrm{\pi } \text{and} \underset{\mathrm{x}\to -\infty }{\lim }\mathrm{f}\left(\mathrm{x}\right)=-\mathrm{\pi }$

Hence, $\mathrm{f}\left(\mathrm{x}\right)$ is monotonically increasing for $\mathrm{x}\in \mathrm{R}$ and the range of $\mathrm{f}\left(\mathrm{x}\right)$ is $\left(-\mathrm{\pi },\mathrm{\pi }\right)$.

Hence, $\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{\pi }}{4}$ has exactly one solution for $\mathrm{x}>0$.