Answer
The identity is verified.
$arctan~x+arctan\frac{1}{x}=\frac{\pi}{2}$
Work Step by Step
Let $y=arctan~x$
$x=tan~y$
$\frac{1}{x}=cot~y=tan(\frac{\pi}{2}-y)$
$arctan~x+arctan\frac{1}{x}=arctan~x+arctan[tan(\frac{\pi}{2}-y)]=y+\frac{\pi}{2}-y=\frac{\pi}{2}$
