If $\alpha, \beta, \gamma$ are the roots of equations $x³+px²+qx+p=0$, Then the value of $\tan^{-1}\alpha+\tan^{-1}\beta+\tan^{-1}\gamma$ is
- A. $n\pi/2$
- B. $n\pi$
- C. $2n\pi$
- D. $n\pi/4$
Answer: B. $n\pi$
Explanation: $\tan^{-1}\alpha+\tan^{-1}\beta+\tan^{-1}\gamma = \tan^{-1}(\frac{\alpha+\beta+\gamma-\alpha\beta\gamma}{1-(\alpha\beta+\beta\gamma+\gamma\alpha)})$. For $x³+px²+qx+p=0$, sum of roots $\alpha+\beta+\gamma = -p$, product of roots $\alpha\beta\gamma = -p$, and $\alpha\beta+\beta\gamma+\gamma\alpha = q$. Thus $\tan^{-1}(\frac{-p-(-p)}{1-q}) = \tan^{-1}(0) = n\pi$.