Question

If $x_1,x_2,x_{3}and{x}_4$ are the roots of the equation

$x^4-x^{3}sin2\beta +x^{2}cos2\beta -xcos\beta -sin\beta =0$,then

$ta{n}^{-1}{x}_1+ta{n}^{-1}{x}_2+ta{n}^{-1}{x}_3+ta{n}^{-1}{x}_4:$will be equal to

• A. $n\pi +\left(\pi /2\right)-\beta$,where $n$ is an integer.
• B. $n\pi -\beta$,where $n$ is an integer.
• C. $n\pi +\left(\pi /2\right)-2\beta$,where $n$ is an integer.
• D. $2n\pi +\left(\pi /2\right)-\beta$,where $n$ is an integer.

Answer 1

The correct option is A $n\pi +\left(\pi /2\right)-\beta$,where $n$ is an integer.

We have

$S_1=\sum x_1=\sin 2\beta$

$S_2=\sum x_1{x}_2=\cos 2\beta$

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

$S_4=x_1{x}_2{x}_3{x}_4=-\sin \beta$

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

$={\tan }^{-1}\frac{\sin 2\beta -\cos \beta }{1-\cos 2\beta -\sin \beta }$

$={\tan }^{-1}\frac{\cos \beta \left(2\sin \beta -1\right)}{\sin \beta \left(2\sin \beta -1\right)}$

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