If $A$, $B$, and $C$ are the roots of the equation $x^3+mx^2+3x+m=0$, then the value of $$\arctan(A)+\arctan(B)+\arctan(C)$$is given by

A. $n\pi$
B. $n\pi/2$
C. $\pi(2n+1)/2$
D. none

I met this question in JEE Exams of GOIIT but my human ability couldn't solve and I messed up on this question despite being a multiple choice question.

  • $\begingroup$ I've improved your question's formatting; apologies if I changed your meaning. You can see here how I edited your question. Please see here for a guide to writing math with MathJax, and see here for a guide to formatting posts with Markdown. $\endgroup$ Jul 8, 2013 at 23:33
  • $\begingroup$ @zev chonoles thanks but nothing was change in your edit $\endgroup$ Jul 8, 2013 at 23:36
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1 Answer 1


First, prove that

$$\arctan{A}+\arctan{B}+\arctan{C} = \arctan{\frac{A+B+C-ABC}{1-(AB+AC+BC)}}$$

Do this by observing that

$$\arctan{X}+\arctan{Y} = \arctan{\frac{X+Y}{1-XY}}$$

Then note that

$$A+B+C = ABC = -m$$

Thus the sum of the arctangents of the roots is

$$\arctan{0} = n \pi $$

for some $n \in \mathbb{Z}$


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