Answer
See explanations.
Work Step by Step
Step 1. Let $u=tan^{-1}x$; we have $tan(u)=x$. As $x\gt0$, we know $u\in(0,\frac{\pi}{2})$
Step 2. Form a right triangle with opposite side=$x$, adjacent side=$1$, As $tan(u)=x$, we know one acute angle is $u$ and the other is $v=\frac{\pi}{2}-u$ or $u+v=\frac{\pi}{2}$
Step 3. In the same triangle, we have $tan(v)=\frac{1}{x}$; thus $v=tan^{-1}\frac{1}{x}$
Step 4. Combining the above results, we have $u+v=tan^{-1}x+tan^{-1}\frac{1}{x}=\frac{\pi}{2}$
