Verify the identity:
arctanx+ arctan (1/x)=( ∏/2)
using calculus theory.
differentiate the left hand side of the identity.
If $\displaystyle f(x)=\arctan x+\arctan (1/x)\; (x\neq 0)$ then, $\displaystyle f'(x)=\ldots=0$ . As $\displaystyle (0,+\infty)$ is connected, then $\displaystyle f(x)$ is constant in $\displaystyle (0,+\infty)$. But $\displaystyle f(1)=\pi/4+\pi/4=\pi/2$ so $\displaystyle f(x)=\pi/2$ in $\displaystyle (0,+\infty)$ . Similar arguments for $\displaystyle (-\infty,0)$ .
Take as a counterexample:
$\displaystyle f(x)=\begin{Bmatrix} 1 & \mbox{ if }& x<0\\0 & \mbox{if}& x>0\end{matrix}$
Then, $\displaystyle f'(x)=0$ for all $\displaystyle x$ on its domain and $\displaystyle f(x)$ is not constant. As I said in answer #2 the domain must be connected, or equivalently (in $\displaystyle \mathbb{R}$) an interval.