# Sum of the arctangents of roots of a cubic equation (multiple choice)

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.

• 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. Jul 8, 2013 at 23:33
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– user17762
Dec 6, 2013 at 7:15

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 $n \in \mathbb{Z}$