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. $-\beta$

Answer 1

The correct option is B

$\frac{\pi }{2}-\beta$

We have $\sum x_1=sin2\beta ,\sum x_1{x}_2=cos2\beta \sum x_1{x}_2{x}_3=cos\beta and{x}_1.x_2.x_3.x_4=-sin\beta \therefore ta{n}^{-1}{x}_1+ta{n}^{-1}{x}_2+ta{n}^{-1}{x}_3+ta{n}^{-1}{x}_4=ta{n}^{-1}\left(\frac{\sum x_1-\sum x_1{x}_2{x}_3}{1-\sum x_1{x}_2+x_1{x}_2{x}_3{x}_4}\right)=ta{n}^{-1}\left(\frac{sin2\beta -cos\beta }{1-cos2\beta -sin\beta }\right)=ta{n}^{-1}\left(\frac{(2sin\beta -1)cos\beta }{sin\beta (2sin\beta -1)}\right)=ta{n}^{-1}\left[tan(\frac{\pi }{2}-\beta )\right]=\frac{\pi }{2}-\beta$