If $x_{1}, x_{2}, x_{3}, x_{4}$ are roots of 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 }=$

Let the roots be x_i, and consider the tangent of the sum of the arctangent angles, tan(sum of arctan(x_i)), which expands as (S1 - S3) / (1 - S2 + S4) using the tangent addition formula. From the given equation, the sums of roots are S1 = sin(2b), S2 = cos(2b), S3 = -cos(b), and S4 = -sin(b). Substituting these values yields (sin(2b) + cos(b)) / (1 - cos(2b) - sin(b)), which simplifies to cos(b) / sin(b) = cot(b). Taking the inverse tangent gives arctan(cot(b)), which equals pi/2 - b. The sum of the inverse tangents is pi/2 - b.

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