Q. If $\alpha,\beta$ are the roots of the equation $6x^{2} - 5x + 1 = 0$, then the value of $\tan^{- 1}{\alpha + \tan^{- 1}\beta}$ is
A) 0
B) $\pi/4$
C) 1
D) $\pi/2$
Explanation
(b)
Here, $\alpha + \beta = \frac{5}{6}$ and $\alpha\beta = \frac{1}{6}$
$\therefore\ \tan^{- 1}{\alpha + \tan^{1}{\beta = \tan^{- 1}\left( \frac{\alpha + \beta}{1 - \alpha\beta} \right)}}$
$= \tan^{- 1}\left( \frac{\frac{5}{6}}{1 - \frac{1}{6}} \right)$
$= \tan^{- 1}{1 = \frac{\pi}{4}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.