Let, \quad tan(y)=x=x/1
Because tan(\text{angle})=\text{opposite}/\text{adjacent}, we have the diagram below:
The upper angle is \pi/2-y because the sum of the angles of a triangle needs to be \pi\quad (That is, 180 degrees).
Also note from the diagram that since cot(\text{angle})=\text{adjacent}/\text{opposite}
we have that cot(\pi/2-y)=x/1.
Invert the two expressions for tan and cot.
cot^{-1}(x)=\pi/2-y
tan^{-1}(x)=y
Therefore,
cot^{-1}(x)+tan^{-1}(x)=(\pi/2-y)+(y)=\pi/2
Note that this proof requires that we use the reference angles when taking inverses, the tan function is periodic so
tan(y+2\pi)=tan(y)=x
so we could take tan^{-1}(x) and get y+2pi and this would make the identity not true. But if we take the smallest reference angles, the identity is true.