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 ${\tan }^{-1}{x}_1+{\tan }^{-1}{x}_2+{\tan }^{-1}{x}_3+{\tan }^{-1}{x}_4=$

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

Answer 1

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

From the given equation,
$\sum x_1=\sin 2\beta ;\sum x_1{x}_2=\cos 2\beta ;\sum x_1{x}_2{x}_3=\cos \beta$
and $x_1{x}_2{x}_3{x}_4=-\sin \beta$
$\therefore {\tan }^{-1}{x}_1+{\tan }^{-1}{x}_2+{\tan }^{-1}{x}_3+{\tan }^{-1}{x}_4=$
$={\tan }^{-1}\left[\begin{array}{c}\frac{\sum x_1-\sum x_1{x}_2{x}_3}{1-\sum x_1{x}_2+x_1{x}_2{x}_3{x}_4}\end{array}\right]$
$={\tan }^{-1}\left[\begin{array}{c}\frac{\sin 2\beta -\cos \beta }{1-\cos 2\beta -\sin \beta }\end{array}\right]$
$={\tan }^{-1}\left[\begin{array}{c}\frac{(2\sin \beta -1)\cos \beta }{\sin \beta (2\sin \beta -1)}\end{array}\right]={\tan }^{-1}\left(\cot \beta \right)$
$={\tan }^{-1}\left[\begin{array}{c}\tan (\frac{\pi }{2}-\beta )\end{array}\right]=\frac{\pi }{2}-\beta$