Roots of a $5$th degree polynomial are $\tan A, \tan B, \ldots, \tan E$ If $\tan A$, $\tan B$, $\tan C$, $\tan D$, and $\tan E$ are the roots of the equation $x^5-3x^4-4x^2+3x-5=0$, find the sum of all principal values of $A+B+C+D+E$. 
What do I do with that equation? Thanks.

Hint: let $s_1$ through $s_5$ denote the elementary symmetric polynomials in 5 variables. Set $x_1=\tan A$, $x_2=\tan B$, etc, and try to prove
$$\tan\left(A+B+C+D+E\right)=\frac{s_1(x_1,x_2,x_3,x_4,x_5)-s_3(x_1,x_2,x_3,x_4,x_5)+s_5(x_1,x_2,x_3,x_4,x_5)}{1-s_2(x_1,x_2,x_3,x_4,x_5)+s_4(x_1,x_2,x_3,x_4,x_5)}$$
Then use Vieta's formulas.
