If ${x}{1},{x}{2},{x}{3},{x}{4}$ are the roots of the equation ${x}^{4}-{x}^{3}\sin{2\beta}+{x}^{2}\cos{2\beta}-x\cos{\beta}-\sin{\beta}=0$, then $\tan ^{ -1 }{ { x }{ 1 } } +\tan ^{ -1 }{ { x }{ 2 } } +\tan ^{ -1 }{ { x }{ 3 } } +\tan ^{ -1 }{ { x }{ 4 } } =n\pi +\cfrac { \pi }{ 2 } -\beta $.

Let S be the sum of the inverse tangents of the roots x1, x2, x3, x4. Taking the tangent of both sides, tan(S) equals the sum of the roots minus the sum of products of roots taken three at a time, all divided by 1 minus the sum of products taken two at a time plus the product of all roots. Using Vieta's formulas for the given quartic, the numerator is sin(2b) - cos(b) and the denominator is 1 - cos(2b) - sin(b); this fraction exactly simplifies to cot(b) using trigonometric identities. Therefore, tan(S) = cot(b) = tan(pi/2 - b), which implies the sum evaluates to n*pi + pi/2 - b, making the statement true.

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