Answer
See below.
Work Step by Step
We consider only the case with $v\ge0$.
Step 1. Letting $tan^{-1}v=x$, we have $tan(x)=v$ where $x\in[0,\frac{\pi}{2}]$
Step 2. Letting $cot^{-1}v=y$, we have $cot(y)=v$ where $y\in[0,\frac{\pi}{2}]$
Step 3. We have $tan(x)=cot(y)=tan(\frac{\pi}{2}-y)$, thus $x=k\pi+\frac{\pi}{2}-y$ and $x+y=k\pi+\frac{\pi}{2}$
Step 4. As $(x+y)\in [0,\pi]$, we have $k=0$ and $x+y=\frac{\pi}{2}$
