Prove that arctan (2+sqrt3)+arctan (2-sqrt3)=pi/2!

The terms arctan (2+sqrt3) and arctan (2-sqrt3) are
angles.

Let's note arctan (2+sqrt3) = x and arctan
(2-sqrt3) = y

We'll apply the tangent function to the
addition of angles x and y:

tan (x+y) = (tan x + tany)/(1 -
tan x*tany)

where:

tan x = tan
[arctan (2+sqrt3)] = 2+sqrt3

tan y = tan [arctan (2-sqrt3)]
= 2-sqrt3

tan x*tan y = (2+sqrt3)(2-sqrt3) = 4-3 =
1

Let's substitute the values into the
formula:

tan (x+y) = (2+sqrt3+2-sqrt3) / (1
-  1)

We'll eliminate like
terms:

tan (x+y) = 4/0

tan
(x+y) = infinite => x+y = pi/2

arctan
(2+sqrt3)+arctan (2-sqrt3)=pi/2 q.e.d.