Question

if x1, x2, x3, x4 are the roots of the equation
$x^{4} - x^{3}sin 2\beta + x^{2} cos2\beta -xcos\beta - sin \beta =0$
then find the value of
$tan^{-1} x1+tan^{-1}x2+tan^{-1}x3+tan^{-1}x4$
and your option are following
a) β
b)(π/2)-β
c)π-β
d) -β
as i think gys it will solve by attached image

Answer 1

Solution:

$x_{1},x_{2},x_{3},x_{4}$ are the roots of the equation ,which is

$x^{4} - x^{3}sin 2\beta + x^{2} cos2\beta -xcos\beta - sin \beta =0$

                                                                                                ------------------------(1)

For, a fourth root Polynomial

$ax^4+bx^3+cx^2+dx+e=0$

,having roots ,p,q,r and s,the relationship between root and polynomial can be elaborated in the following way:

$p+q+r+s=\frac{-b}{a}\\\\pq+qr+rs+sp=\frac{c}{a}\\\\pqr+qrs+srp=\frac{-d}{a}\\\\pqrs=\frac{e}{a}$

The roots of the equation 1,can be written in following way

$1.x_{1}+x_{2}+x_{3}+x_{4}=sin 2\beta\\\\2.x_{1}x_{2}+x_{2}x_{3}+x_{3}x_{4}+x_{4}x_{1}=cos2\beta\\\\3.x_{1}x_{2}x_{3}+x_{2}x_{3}x_{4}+x_{2}x_{3}x_{1}=cos\beta\\\\4.x_{1}x_{2}x_{3}x_{4}=- sin \beta$

$tan^{-1} x_{1}+tan^{-1}x_{2}+tan^{-1}x_{3}+tan^{-1}x_{4}=tan^{-1}[\frac{(x_{1}+x_{2}+x_{3}+x_{4})-(x_{1}x_{2}x_{3}+x_{2}x_{3}x_{4}+x_{2}x_{3}x_{1})}{1-(x_{1}x_{2}+x_{2}x_{3}+x_{3}x_{4}+x_{4}x_{1})-x_{1}x_{2}x_{3}x_{4}}]\\\\tan^{-1}[\frac{sin2\beta-cos\beta}{1-cos2\beta-sin\beta}]\\\\=tan^{-1}[\frac{2\times sin\beta\times cos\beta-cos\beta}{1-1+2sin^2\beta-sin\beta}]\\\\=tan^{-1}[\frac{cos\beta(2sin\beta-1)}{sin\beta(2sin\beta-1)}]\\\\=tan^{-1}[cot\beta]\\\\=tan^{-1}[tan(\frac{\pi}{2}-\beta}]$

  $=\frac{\pi}{2}-\beta$

Option B  

Used the trigonometric identity

Sin2A=2 SinA×CosA 

Cos2A=2Cos²A-1=1-2sin²A=Cos²A-Sin²A