Question:

<div class="solution"><p><strong>Step 1:</strong> Roots are 1 and 2. <span class="math-inline">$\alpha\beta=2>1$</span>.</p><p><strong>Step 2:</strong> Since <span class="math-inline">$\alpha\beta>1$</span> and both positive: <span class="math-block">$$\tan^{-1}1+\tan^{-1}2=\pi+\tan^{-1}\!\frac{3}{1-2}=\pi-\tan^{-1}3$$</span></p><p><strong>Answer: (2) <span class="math-inline">$\pi-\tan^{-1}3$</span></strong></p><div class="trap-box"><strong>Trap:</strong> Raw formula gives tan⁻¹(-3), missing the π quadrant correction. Sum of two positive acute angles must lie in (π/2,π).</div><div class="key-concept"><strong>Key Concept:</strong> For tan⁻¹a+tan⁻¹b: if a,b>0 and ab>1, add π to the formula result</div></div>