Question

lf $\alpha ,\beta ,\gamma$ are the roots of the equation $x^3+m{x}^2+3x+m=0$, then the general value of ${\tan }^{-1}\alpha +{\tan }^{-1}\beta +{\tan }^{-1}\gamma$ is:

• A. $(2n+1)\frac{\pi }{2}$
• B. $n\pi$
• C. $\frac{n\pi }{2}$
• D. dependent upon the value of m

Answer 1

The correct option is B $n\pi$

$x^3+m{x}^2+3x+m=0$ have roots $\alpha ,\beta ,\gamma$
$\alpha +\beta +\gamma =-m$
$\alpha \beta +\beta \gamma +\alpha \gamma =3$
$\alpha \beta \gamma =-m$
$\therefore ta{n}^{-1}\alpha +ta{n}^{-1}\beta +ta{n}^{-1}\gamma =ta{n}^{-1}\left(\frac{\alpha +\beta }{1-\alpha \beta }\right)+ta{n}^{-1}\left(\gamma \right)$
$=ta{n}^{-1}\left(\frac{\frac{\alpha +\beta }{1-\alpha \beta }+\gamma }{1-\gamma \left(\frac{\alpha +\beta }{1-\alpha \beta }\right)}\right)$
$=ta{n}^{-1}\left(\frac{\alpha +\beta +\gamma -\alpha \beta \gamma }{1-\alpha \beta -\beta \gamma -\alpha \gamma }\right)$
$=ta{n}^{-1}\left(\frac{-m+m}{1-3}\right)$
$=ta{n}^{-1}\left(\frac{m-m}{-2}\right)=ta{n}^{-1}\left(0\right)=\pi$
and general solution is $n\pi$