First slide

Properties of ITF

Question

If $x_{1}, x_{2}, x_{3}, x_{4}$ are roots of the equation $x^{4} - x^{3} \sin 2 β + x^{2} \cos 2 β - x \cos β - \sin β = 0$ then ${Tan}^{- 1} x_{1} + {Tan}^{- 1} x_{2} + {Tan}^{- 1} x_{3} + {Tan}^{- 1} x_{4} =$

Moderate

A

$β$
B

$\frac{π}{2} - β$
C

$π - β$
D

$- β$

Solution

from the given equation
$\sum x_{1} = \sin 2 β;$ $\sum x_{1} x_{2} = \cos 2 β;$ $\sum x_{1} x_{2} x_{3} = \cos β;$
and $x_{1} x_{2} x_{3} x_{4} = - \sin β$
$\begin{array}{l} ∴ \tan^{- 1} (x_{1}) + \tan^{- 1} x_{2} + \tan^{- 1} x_{3} + \tan^{- 1} x_{4} \\ = \tan^{- 1} [\frac{\sum x_{1} - \sum x_{1} x_{2} x_{3}}{1 - \sum x_{1} x_{2} + x_{1} x_{2} x_{3} x_{4}}] \\ = \tan^{- 1} (\frac{\sin 2 β - \cos β}{1 - \cos 2 β - \sin β}) \\ = \tan^{- 1} (\frac{2 \sin β \cos β - \cos β}{2 \sin^{2} β - \sin β}) \\ = \tan^{- 1} (\frac{\cos β (2 \sin β - 1)}{\sin β (2 \sin β - 1)}) \\ = \tan^{- 1} (c o t β) \\ = \tan^{- 1} (\tan (\frac{π}{2} - β)) \\ = \frac{π}{2} - β \end{array}$

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