$\alpha ,\beta$, and  $\gamma$ are the roots of $x(x-200)(4x+1)=1$.
Let  $\omega ={\text{tan}}^{-1}\left(\alpha \right)+{\text{tan}}^{-1}\left(\beta \right)+{\text{tan}}^{-1}\left(\gamma \right).$ The value of $\text{tan}\left(\omega \right)$ can be written as $\frac{m}{n}$ where m  and n are relatively prime positive integers. Determine the value of $m+n$. (where ${\tan }^{-1}x\in (-\infty ,\infty )$  & $x\in \left(-\frac{\pi }{2},\frac{\pi }{2}\right)$ )

We know that  $\alpha ,\beta ,\gamma$ are the roots of
  $x(x-200)(x+1/4)-1/4=x^3-\frac{799}{4}{x}^2-50x-\frac{1}{4}$
By Vieta's formulas, we have:
$\begin{array}{rr} & \alpha +\beta +\gamma =\frac{799}{4}\\ & \alpha \beta +\beta \gamma +\gamma \alpha =-50\\ & \alpha \beta \gamma =\frac{1}{4}\end{array}$

Now, by tangent addition formulas, we have $\text{tan}\left(\omega \right)=\frac{\alpha +\beta +\gamma -\alpha \beta \gamma }{1-\alpha \beta -\beta \gamma -\gamma \alpha }$ . Substituting
Vieta's formulas, we obtain
$\text{tan}\left(\omega \right)=\frac{\frac{799}{4}-\frac{1}{4}}{1-(-50)}=\frac{\frac{798}{4}}{51}=\frac{133}{34}\text{.}$
Therefore, our answer is $133+34=167$  and we are done.