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.