If $x_{1}, x_{2}, x_{3}, x_{4}$ are the roots of the equation $x_{4}- x_{3}$ $\sin 2\beta + x^{2} \cos2\beta - x\cos \beta -\sin \beta=0 B \epsilon (0,\pi)$ , where $\sum_{i=1}^{4}\tan^{-1} X_{i}$ is equal to

Let S = sum arctan(xi). By taking the tangent of both sides and using the addition formula for four angles, tan(S) = (Sum xi - Sum of products of 3 roots) / (1 - Sum of products of 2 roots + Product of 4 roots). By Vieta's formulas, the numerator is sin(2b) - cos(b) and the denominator is 1 - cos(2b) - sin(b), which simplifies to cot(b). Therefore, tan(S) = cot(b) = tan(pi/2 - b), meaning S = pi/2 - b.

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