If tan-1- tan-2- tan-3 and tan-4 are the roots of the equation x4-x3sin 2 -x2cos 2 -x cos-sin-0 then tan-1-2-3-4-

If $t a n θ_{1}, t a n θ_{2}, t a n θ_{3}$ are the real roots of $x^{3} - (a + 1) x^{2} + (b - a) x - b = 0 w h e r e θ_{1}, θ_{2}, θ_{3}$ are acute then $θ_{1} + θ_{2} + θ_{3} =$

Q. 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} =

Q. 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

4 \sum i = 1 {tan}^{- 1} x_{i}

is equal to

(

where

β \in (0, \frac{π}{2}) - {\frac{π}{6}})

Join BYJU'S Learning Program

If tanθ1,tanθ2,tanθ3 and tanθ4 are the roots of the equation x4−x3 sin2β+x2cos2β−xcosβ−sinβ=0 then tan(θ1+θ2+θ3+θ4)=

The correct option is A cotβtan(θ1+θ2+θ3+θ4)=S1−S31−S2+S4 =sin2β−cosβ1−cos2β−sinβ=cosβ(2sinβ−1)2sin2β−sinβ=cotβ

If $t a n θ_{1}, t a n θ_{2}, t a n θ_{3}$ and $t a n θ_{4}$ are the roots of the equation $x^{4} - x^{3} s i n 2 β + x^{2} c o s 2 β - x c o s β - s i n β = 0$ then $t a n (θ_{1} + θ_{2} + θ_{3} + θ_{4}) =$

The correct option is A $c o t β$
$t a n (θ_{1} + θ_{2} + θ_{3} + θ_{4}) = \frac{S_{1} - S_{3}}{1 - S_{2} + S_{4}}$
$= \frac{s i n 2 β - c o s β}{1 - c o s 2 β - s i n β} = \frac{c o s β (2 s i n β - 1)}{2 s i n^{2} β - s i n β} = c o t β$