Question:

<div class="solution"><p><strong>Step 1:</strong> Expand: <span class="math-inline">$3x^3-13x^2+14x-2=0$</span>.</p><p><strong>Step 2:</strong> Vieta's: <span class="math-inline">$S_1=13/3,\;S_2=14/3,\;S_3=2/3$</span>.</p><p><strong>Step 3:</strong> <span class="math-block">$$\tan(\tan^{-1}r+\tan^{-1}s+\tan^{-1}t)=\frac{S_1-S_3}{1-S_2}=\frac{13/3-2/3}{1-14/3}=\frac{11/3}{-11/3}=-1$$</span></p><p><strong>Step 4:</strong> Since <span class="math-inline">$r,s,t>0$</span>, each <span class="math-inline">$\tan^{-1}\in(0,\pi/2)$</span>, so sum <span class="math-inline">$\in(0,3\pi/2)$</span>. Only <span class="math-inline">$3\pi/4$</span> gives <span class="math-inline">$\tan=-1$</span> there.</p><p><strong>Answer: (B) <span class="math-inline">$3\pi/4$</span></strong></p><div class="trap-box"><strong>Trap:</strong> tan = -1 also at -π/4, but sum of positive arc-tangents must be positive.</div><div class="key-concept"><strong>Key Concept:</strong> Vieta + tan⁻¹ addition formula + quadrant check for the sum</div></div>