Question

If $\alpha ,\beta ,\gamma$ be the roots of equation $x^3+p{x}^2+qx+p=0$, then prove that, except a special condition,
${\tan }^{-1}\alpha +{\tan }^{-1}\beta +{\tan }^{-1}\gamma =n\pi$
also find the special condition, when it does no so.

Answer 1

${\tan }^{-1}\alpha +{\tan }^{-1}\beta +{\tan }^{-1}{\gamma }^2=n\pi$

$\Rightarrow {\tan }^{-1}\left[\frac{\alpha +\beta +\gamma -\alpha \beta \gamma }{1-\sum \alpha \beta }\right]=n\pi$

$\therefore \frac{\alpha +\beta +{\gamma }^2-\alpha \beta \gamma }{1-\sum \alpha \beta }=0$

$\Rightarrow \alpha +\beta +{\gamma }^2=\alpha \beta \gamma$ ………$\left(1\right)$

$\sum \alpha \beta \ne 1$ …….$\left(2\right)$

comparing with equation

$x^3+p{x}^2+qx+p=0$

$\alpha +\beta +\gamma =-p$

$\alpha \beta \gamma =-p$

and $\sum \alpha \beta =q\ne 1$

are the conditions if $\alpha ,\beta ,\gamma$ are the roots of equation.