Answer
See below.
Work Step by Step
1. Let $tan^{-1}v=x$, we have $tan(x)=v, sin(x)=\frac{v}{\sqrt {1+v^2}}, cos(x)=\frac{1}{\sqrt {1+v^2}}$
2. Let $cot^{-1}v=y$, we have $cot(y)=v, sin(y)=\frac{1}{\sqrt {1+v^2}}, cos(y)=\frac{v}{\sqrt {1+v^2}}$
3. $sin(x+y)=sin(x)cos(y)+cos(x)sin(y)=(\frac{v}{\sqrt {1+v^2}})(\frac{v}{\sqrt {1+v^2}})+(\frac{1}{\sqrt {1+v^2}})(\frac{1}{\sqrt {1+v^2}})=1$
4. Thus $x+y=2k\pi+\frac{\pi}{2}$
5. Consider the ranges of $x,y$, we have $x+y=\frac{\pi}{2}$
