If  $x_1,x_2,x_3,x_4$  are the roots of the equation $x^4-x^3\sin 2\beta +x^2\cos 2\beta -x\cos \beta -\sin \beta =$,  then the value of $\sum _{i=1}^4{\tan }^{-1}{x}_i$ is

If  $x_1,x_2,x_3,x_4$  are the roots of the equation

$\begin{array}{l}{x}^4-x^3\sin 2\beta +x^2\cos 2\beta -x\cos \beta -\sin \beta =\\ \sum x_1=\sin 2\beta ,\\ \sum x_1{x}_2=\cos 2\beta ,\\ \sum x_1{x}_2{x}_3=\cos \beta ,\\ \sum x_1{x}_2{x}_3{x}_4=-\sin \beta ,\end{array}$

$\begin{array}{l}{\tan }^{-1}{x}_1+{\tan }^{-1}{x}_2+{\tan }^{-1}{x}_3+{\tan }^{-1}{x}_4={\tan }^{-1}\frac{{\displaystyle \sum x_1-\sum x_1{x}_2{x}_3}}{1-[x_1{x}_2+x_1{x}_2{x}_3{x}_4]}\\ ={\tan }^{-1}\left(\frac{\sin 2\beta -\cos \beta }{1-\cos 2\beta -\sin \beta }\right)\\ ={\tan }^{-1}\left(\frac{(2\sin 2\beta -1)\cos \beta }{(2\sin 2\beta -1)\sin \beta }\right)\end{array}$

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