If $\alpha, \beta, \gamma$ are the roots of the equation $$x^3 + px^2 + qx + p = 0,$$
prove that $$\tan^{-1}\left(\alpha\right) + \tan^{-1}\left(\beta\right) + \tan^{-1}\left(\gamma\right) = n\pi$$
except in one particular case.
This question is from S. L. Loney's 'Plane Trigonometry' page 327 q13.
It may be useful to note that this section utilises $$\tan\left(\alpha + \beta + \gamma + …\right) = \frac{s_1 – s_3 + s_5}{1 – s_2 + s_4 – \cdots}$$
where
$s_1 =$ the sum of the tangents of the separate angles,
$s_2 =$ the sum of the tangents taken two at a time,
$s_3 =$ the sum of the tangents taken three at a time, and so on.
I do not know where I should start with this question.
Best Answer
Let $S=\tan^{-1}(\alpha) + \tan^{-1}(\beta) + \tan^{-1}(\gamma)$. Then, by your formula, $$ \tan S = \frac{s_1 - s_3}{1-s_2}, $$ where $s_1 = \alpha+\beta+\gamma$, $s_2 = \alpha\beta + \beta\gamma + \gamma\alpha$ and $s_3 = \alpha\beta\gamma$. Using Vieta's formulas, $$ s_1 = -p,\,\,\,s_2 = q\,\,\,\text{and }s_3=-p, $$ therefore $\tan S = 0$, so $S=n\pi$ for some $n\in \mathbb{Z}$. The particular case is where $s_2 = q = 1$ (where you would get $0$ in the denominator), and then $\tan S = \pm \infty$, giving $S=n\pi/2$ for $n\in \mathbb{Z}$ not divisible by $2$.