Q. If $x_{1},x_{2},\ x_{3},\ x_{4}$ are the roots of the equation $x^{4} - x^{3}\sin{2\beta - x\cos{\beta - \sin{\beta = 0,}}\ }$ then $\tan^{- 1}{+ \tan^{- 1}{x_{2} + \tan^{- 1}{x_{3} + \tan^{- 1}x_{4}}}}$ is equal to

A) $\beta$
B) $\frac{\pi}{2} - \beta$
C) $\pi - \beta$
D) $- \beta$
Explanation

(b)
We have, $\Sigma x_{1} = \sin{2\beta,\ \Sigma x_{1}x_{2} = \cos{2\beta,\ \Sigma x_{1}x_{2}x_{3} = \cos\beta}}$ and $x_{1}x_{2}x_{3}x_{4} = - \sin\beta$
$\therefore\tan^{- 1}{x_{1} + \tan^{- 1}{x_{2} + \tan^{- 1}{x_{3} + \tan^{- 1}x_{4}}}}$
$= \tan^{- 1}\left( \frac{\Sigma x_{1} - \Sigma x_{1}x_{2}x_{3}}{1 - \Sigma x_{1}x_{2} + x_{1}x_{2}x_{3}x_{4}} \right)$
$= \tan^{- 1}\left( \frac{\sin{2\beta - \cos\beta}}{1 - \cos{2\beta - \sin\beta}} \right)$
$= \tan^{- 1}\left( \frac{(2\sin{\beta - 1)\cos\beta}}{\sin{\beta(2\sin{\beta - 1)}}} \right)$
$= \tan^{- 1}{(\cot{\beta)}}$
$= \tan^{- 1}{\left( \tan\left( \frac{\pi}{2} - \beta \right) \right) = \frac{\pi}{2} - \beta}$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.