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.
