Answer
See below.
Work Step by Step
Step 1. Letting $tan^{-1}v=x$, we have $tan(x)=v$ with $x\in(0,\frac{\pi}{2}]$
Step 2. Letting $tan^{-1}\frac{1}{v}=y$, we have $tan(y)=\frac{1}{v}$ $y\in(0,\frac{\pi}{2}]$
Step 3. We have $tan(x)=\frac{1}{tan(y)}=cot(y)=tan(\frac{\pi}{2}-y)$, 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 $x+y=\frac{\pi}{2}$ or $y=\frac{\pi}{2}-x$
