Question

If $x_1,x_2,x_3,x_4$ are roots of the equation $x^4–x^3$ $sin2\beta +x^{2}cos2\beta –xcos\beta –sin\beta =0then{\sum }_{i=1}^{4}ta{n}^{-1}{x}_i$ is equal to

• A. $\pi -\beta$
• B. $\pi -2\beta$
• C. $\frac{\pi }{2}-\beta$
• D. $\frac{\pi }{2}-2\beta$

Answer 1

The correct option is C

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

We have

$S_1=\sum x_1=sin2\beta$

$S_2=\sum x_1{x}_2=cos2\beta$

$S_3=\sum x_1{x}_2{x}_3=cos\beta$

$S_4=x_1{x}_2{x}_3{x}_4=–sin\beta$

So that ${\sum }_{i=1}^{4}ta{n}^{-1}{x}_i=ta{n}^{-1}\frac{S_1-S_3}{1-S_2+S_4}$

$=ta{n}^{-1}\frac{sin2\beta -cos\beta }{1-cos2\beta -sin\beta }$

$=ta{n}^{-1}\frac{cos\beta (2sin\beta -1)}{sin\beta (2sin\beta -1)}$

$=ta{n}^{-1}cot\beta =ta{n}^{-1}\left(tan\right(\frac{\pi }{2}-\beta \left)\right)$

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