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If $\tan θ_{1}, \tan θ_{2}, \tan θ_{3}$ are the real roots of the equation $x^{3} - (a + 1) x^{2} + (b - a) x - b = 0,$ where $θ_{1} + θ_{2} + θ_{3} \in (0, π)$ , then $θ_{1} + θ_{2} + θ_{3}$ is equal to

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$\frac{π}{2}$

$\frac{π}{4}$

$\frac{3 π}{4}$

$π$

answer is B.

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Detailed Solution

s₁=(a+1) s₂=(b-a) s₃=b tan(tita1+tita2+tita3)=1

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