This question already has an answer here:

Anyone know how to force Mathematica to return:


I've tried this:

ArcTan[x] + ArcTan[1/x] // FullSimplify


Pi/2 - ArcTan[1/x] // FullSimplify

and several other methods, but no luck thus far.


marked as duplicate by Jens, Bob Hanlon, Simon Woods, corey979, BlacKow Jan 4 '17 at 22:07

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  • 1
    $\begingroup$ Note that x needs be positive $\endgroup$ – Feyre Jan 4 '17 at 19:28
  • 1
    $\begingroup$ This doe not solve your problem, but if you plot ArcTan[x]+ArcTan[1/x] you see that your expression is valid only for x>=0 $\endgroup$ – mattiav27 Jan 4 '17 at 19:31
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    $\begingroup$ Assuming[x > 0, ArcTan[x] + ArcTan[1/x] // TrigToExp // FullSimplify] $\endgroup$ – Bob Hanlon Jan 4 '17 at 19:32
  • $\begingroup$ These comments are so helpful. Thanks. $\endgroup$ – David Jan 4 '17 at 22:14

This might not be elegant (but may be Bob's method is simpler in the comment)

 expr = ArcTan[x] + ArcTan[1/x];
 Normal@Assuming[x > 0, Series[expr, {x, 0, 5}]]

Mathematica graphics


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