Question:

<div class="solution"><p><strong>Step 1:</strong> $\cot^{-1}9=\tan^{-1}(1/9)$.</p><p><strong>Step 2:</strong> $\csc\theta=\sqrt{41}/4\implies\sin\theta=4/\sqrt{41}$, adjacent $=5$, so $\tan\theta=4/5\implies\theta=\tan^{-1}(4/5)$.</p><p><strong>Step 3:</strong> <span class="math-block">$$\tan^{-1}(1/9)+\tan^{-1}(4/5)=\tan^{-1}\!\frac{1/9+4/5}{1-4/45}=\tan^{-1}\!\frac{41/45}{41/45}=\tan^{-1}1=\pi/4$$</p><p><strong>Answer: (2) \pi/4</strong></p><div class="trap-box"><strong>Trap:</strong> Keep cosec⁻^1 in its original form -- convert to tan⁻^1 via right triangle first.<div class="key-concept"><strong>Key Concept:</strong> Convert reciprocal inverse trig via right triangle, then use addition formula