Answer
See below.
Work Step by Step
Step 1. Letting $cot^{-1}e^v=x$, we have $cot(x)=e^v$ with $x\in(0,\frac{\pi}{2}]$
Step 2. Letting $tan^{-1}e^{-v}=y$, we have $tan(y)=e^{-v}$ with $y\in(0,\frac{\pi}{2}]$
Step 3. We have $cot(x)tan(y)=e^0=1$, thus $tan(x)=tan(y)$ and $x=k\pi+y$
Step 4. Considering the intervals of $x,y$ above, we have $k=0$ and $x=y$
