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

$π / 2 - β$
C

$π - β$
D

$- β$

Solution

We have,
$Σ x_{1} = \sin 2 β, Σ x_{1} x_{2} = \cos 2 β,$
$\sum x_{1} x_{2} x_{3} = \cos β$ and $x_{1} x_{2} x_{3} x_{4} = - \sin β .$
$\begin{array}{l} ∴ \tan^{- 1} x_{1} + \tan^{- 1} x_{2} + \tan^{- 1} x_{3} + \tan^{- 1} x_{4} \\ = \tan^{- 1} \{\frac{\sum x_{1} - \sum x_{1} x_{2} x_{3}}{1 - \sum x_{1} x_{2} + x_{1} x_{2} x_{3} x_{4}}\} \end{array}$
$\begin{array}{l} = \tan^{- 1} \{\frac{\sin 2 β - \cos β}{1 - \cos 2 β - \sin β}\} \\ = \tan^{- 1} \{\frac{(2 \sin β - 1) \cos β}{\sin β (2 \sin β - 1)}\} \\ = \tan^{- 1} (\cot β) = \tan^{- 1} \{\tan (\frac{π}{2} - β)\} = \frac{π}{2} - β \end{array}$

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