If {x_1}, {x_2}, {x_3}, {x_4} are roots of th | vedclass

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(B) Given the equation ${x^4} - {x^3}\sin 2\beta + {x^2}\cos 2\beta - x\cos \beta - \sin \beta = 0$. From Vieta's formulas,we have: $S_1 = \Sigma {x_1

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

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(B) Given the equation ${x^4} - {x^3}\sin 2\beta + {x^2}\cos 2\beta - x\cos \beta - \sin \beta = 0$. From Vieta's formulas,we have: $S_1 = \Sigma {x_1} = \sin 2\beta$ $S_2 = \Sigma {x_1}{x_2} = \cos 2\beta$ $S_3 = \Sigma {x_1}{x_2}{x_3} = \cos \beta$ $S_4 = {x_1}{x_2}{x_3}{x_4} = - \sin \beta$ We know that ${	an ^{ - 1}}{x_1} + {	an ^{ - 1}}{x_2} + {	an ^{ - 1}}{x_3} + {	an ^{ - 1}}{x_4} = {	an ^{ - 1}}\left( {\frac{{S_1 - S_3}}{{1 - S_2 + S_4}}} ight)$. Substituting the values: $= {	an ^{ - 1}}\left( {\frac{{\sin 2\beta - \cos \beta }}{{1 - \cos 2\beta - \sin \beta }}} ight)$ $= {	an ^{ - 1}}\left( {\frac{{2\sin \beta \cos \beta - \cos \beta }}{{1 - (1 - 2\sin^2 \beta) - \sin \beta }}} ight)$ $= {	an ^{ - 1}}\left( {\frac{{\cos \beta (2\sin \beta - 1)}}{{2\sin^2 \beta - \sin \beta }}} ight)$ $= {	an ^{ - 1}}\left( {\frac{{\cos \beta (2\sin \beta - 1)}}{{\sin \beta (2\sin \beta - 1)}}} ight)$ $= {	an ^{ - 1}}(\cot \beta ) = {	an ^{ - 1}}\left[ {	an \left( {\frac{\pi }{2} - \beta } ight)} ight] = \frac{\pi }{2} - \beta $.

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} = $

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