TITLE: Sum of the arctangents of roots of a cubic equation (multiple choice)
QUESTION [5 upvotes]: 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.

REPLY [6 votes]: 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 $n \in \mathbb{Z}$