Evaluate trigonometric function

We know that FullSimplify[ArcTan[k] + ArcTan[1/k], k > 0] does not do it. But by first converting to exponentials, now Mathematica does it

   FullSimplify[ TrigToExp[ArcTan[k] + ArcTan[1/k]] , k > 0]

Gives as output $\frac{\pi}{2}$

This works:

Assuming[k > 0, 
 Solve[FullSimplify[ArcTan[k] + ArcTan[1/k] == x], x]]

(* ==> {{x -> Pi/2}} *)

Here, I equate the expression in the question to a symbol x and ask Mathematica what x is. Inserting an apparently trivial Solve sometimes leads to further simplifications.

In this case, the simplification already occurs in the inner step:

Assuming[k > 0, FullSimplify[ArcTan[k] + ArcTan[1/k] == x]]

(* ==> 2 x == Pi *)

But Solve makes sure that you're getting the result with x on one side of the equation.

Edit: Making it work purely with Simplify:

This is the shortest method I could find:

1/FullSimplify[1/(ArcTan[k] + ArcTan[1/k]), k > 0]

(* ==> Pi/2 *)

So here I just added the operation 1/... to get FullSimplify to do what I want. Then I undo the inverse after the simplification.