Considering only the principal values of inverse trigonometric...

Let, 

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

Now, differentiating the above function we get,

$⇒ {f}^{'}\left(x\right) = \dfrac{1}{1 + {x}^{2}} + \dfrac{2}{1 + 4{x}^{2}}$

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

$⇒ \lim_{x \to \infty}f\left(x\right) = \pi\text{and}\lim_{x \to - \infty}f\left(x\right) = - \pi$

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

Hence, $f\left(x\right)=\dfrac{\pi}{4} has exactly one solution$ for x > 0$.</p>$