Question

If $x_1,x_2,x_3,x_4$ are roots of the equation $x^4-x^{3}sin2\beta +x^{2}cos2\beta -xcos\beta -sin\beta =0$ then
$ta{n}^{-1}{x}_1+ta{n}^{-1}{x}_2+ta{n}^{-1}{x}_3+ta{n}^{-1}{x}_4=$

• A. $\beta$
• B. $\frac{\pi }{2}-\beta$
• C. $\pi -\beta$
• D. None

Answer 1

The correct option is B $\frac{\pi }{2}-\beta$

Here $\sum x_1=sin2\beta ,\sum x_1{x}_2=cos2\beta$
$\sum x_1{x}_2{x}_3=cos\beta ,x_1{x}_2{x}_3{x}_4=-sin\beta$
Now let $ta{n}^{-1}{x}_1={\alpha }_1,ta{n}^{-1}{x}_2={\alpha }_2,ta{n}^{-1}{x}_3={\alpha }_3,ta{n}^{-1}{x}_4={\alpha }_4$
$\Rightarrow x_1=tan{\alpha }_1$ etc.
$\therefore tan({\alpha }_1+{\alpha }_2+{\alpha }_3+{\alpha }_4)=\frac{s_1-s_3}{1-s_2+s_4}$
Now $s_1=\sum tan{\alpha }_1=\sum x_1=sin2\beta$
$s_2=\sum tan{\alpha }_{1}tan{\alpha }_2=\sum x_1{x}_2{x}_3=cos2\beta$
$s_3=\sum tan{\alpha }_{1}tan{\alpha }_{2}tan{\alpha }_3=\sum x_1{x}_2{x}_3=cos2\beta$
$s_4=tan{\alpha }_{1}tan{\alpha }_{2}tan{\alpha }_{3}tan{\alpha }_4$
$=x_1.x_2.x_3.x_4=-sin\beta$
$\therefore tan({\alpha }_1+{\alpha }_2+{\alpha }_3+{\alpha }_4)=\frac{sin2\beta -cos\beta }{1-cos2\beta -sin\beta }$
$=\frac{cos\beta (2sin\beta -1)}{sin\beta (2sin\beta -1)}=cot\beta =tan(\frac{\pi }{2}-\beta )$
$\Rightarrow {\alpha }_1+{\alpha }_2+{\alpha }_3+{\alpha }_4=\frac{\pi }{2}-\beta$
i.e., $ta{n}^{-1}{x}_1+ta{n}^{-1}{x}_2+ta{n}^{-1}{x}_3+ta{n}^{-1}{x}_4=\frac{\pi }{2}-\beta$