Question

If $x_1,x_2,x_3,x_4,$ are roots of the equation $x^4-x^3\sin 2\beta +x^2\cos 2\beta -x\cos \beta -\sin \beta =0$, then $\sum _{i=1}^4{\tan }^{-1}{x}_i$ is equal to $($where $\beta \in (0,\frac{\pi }{2})-\left\{\frac{\pi }{6}\right\})$

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

Answer 1

Answer: (C)

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

If $x_1,x_2,x_3,x_4$ are roots of ...........................
We have $S_1=\sum x_1=\sin 2\beta$
$S_2=\sum x_1{x}_2=\cos 2\beta$
$S_3=\sum x_1{x}_2{x}_3=\cos \beta$
$S_4=\sum x_1{x}_2{x}_3{x}_4=-\sin \beta$
so that ${\sum }_{i=1}^4{\tan }^{-1}{x}_1={\tan }^{-1}\frac{S_1-S_3}{1-S_2+S_4}$
$={\tan }^{-1}\frac{\sin 2\beta -\cos \beta }{1-\cos 2\beta -\sin \beta }={\tan }^{-1}\frac{\cos \beta (2\sin \beta -1)}{\sin \beta (2\sin \beta -1)}$
$={\tan }^{-1}\cot \beta ={\tan }^{-}1\left(\tan \right(\pi /2\beta \left)\right)$
$\pi /2-\beta$