If $\tan \theta_{1},\tan \theta _{2},\tan \theta _{3},\tan \theta _{4}$ 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 $\tan (\theta _1+\theta _{2}+\theta _{3}+\theta _{4})=$

Using Vieta's formulas for the equation x^4 - x^3*sin(2B) + x^2*cos(2B) - x*cos(B) - sin(B) = 0. The sum of roots tan(t1)+tan(t2)+tan(t3)+tan(t4) = sin(2B). The product of roots taken three at a time is cos(B). The product of all four is -sin(B). The formula for tan(t1+t2+t3+t4) involves these symmetric sums, which simplifies to cot(B).