# Evaluate trigonometric function

Can anyone help me to evaluate the trigonometric function below?
Please look that the picture for more details.

Here is the code to evaluate:

ArcTan[k] + ArcTan[1/k]


From trigonometry we know that $tan^{-1}(k)+tan^{-1}(\frac{1}{k})=\frac{\pi}{2}$. However, when I put input that expression into Mathematica I get this:

$$ArcTan[\frac{1}{k}]+ArcTan[k]$$

Can anyone help me to get the result of $\frac{\pi}{2}$?

• The identity does not hold for negative values of k. – Mr.Wizard Dec 24 '16 at 18:40
• So let's assume that k is positive, how to add that condition and simplify to the result of pi/2? – anhnha Dec 24 '16 at 18:41
• I tried FullSimplify[ArcTan[k] + ArcTan[1/k], k > 0] but it did not work. I am still looking at this. – Mr.Wizard Dec 24 '16 at 18:42
• I've got only that FullSimplify[ArcTan[k] + ArcTan[1/k] == Pi/2, k > 0] yields True; but ArcTan[k] + ArcTan[1/k] /. k -> 137 // FullSimplify (or with any other k) gives Pi/2 (or 1.5708 if k is inexact). – corey979 Dec 24 '16 at 18:47

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.

• Really interesting, what is operation 1/...? I couldn't find it from Wolfram Documentation. – anhnha Dec 25 '16 at 1:26
• I just meant the operation of taking the inverse: 1/(ArcTan...) - didn't want to repeat the code. It's not meant to be a secret command with three dots. Although I don't blame you for suspecting as much in a language like Mathematica. – Jens Dec 25 '16 at 1:33
• Well, I knew what you meant by three dots but I didn't notice the inverse inside FullSimplify function so I thought 1/... is as a magic command! – anhnha Dec 25 '16 at 1:51