Topic: Engineering Mathematics
If $\alpha$, $\beta$, $\gamma$ are the roots of equations $x^{3}+px^{2}+qx+p=0$ 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. $\frac{n\pi}{4}$
Answer: B. $n\pi$
Explanation: For the cubic equation $x^{3}+px^{2}+qx+p=0$, the roots $\alpha, \beta, \gamma$ satisfy: $\alpha+\beta+\gamma=-p$, $\alpha\beta+\beta\gamma+\gamma\alpha=q$, and $\alpha\beta\gamma=-p$. Using the tangent addition formula, $\tan^{-1}\alpha+\tan^{-1}\beta+\tan^{-1}\gamma = \tan^{-1}\left(\frac{\alpha+\beta+\gamma-\alpha\beta\gamma}{1-(\alpha\beta+\beta\gamma+\gamma\alpha)}\right)$. Substituting the values: $\tan^{-1}\left(\frac{-p - (-p)}{1-q}\right) = \tan^{-1}(0)$. The general value for $\tan^{-1}(0)$ is $n\pi$.