Question:

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