Question:

<div class="solution"><p><strong>Step 1:</strong> Let \(\cot^{-1}x=\theta\), so \(\cot\theta=x\). Then \(\sin\theta=\dfrac{1}{\sqrt{1+x^2}}\).</p><p><strong>Step 2:</strong> Let \(\tan^{-1}\!\left(\frac{1}{\sqrt{1+x^2}}\right)=\alpha\), so \(\tan\alpha=\frac{1}{\sqrt{1+x^2}}\).</p><p><strong>Step 3:</strong> In the right triangle: opposite=1, adjacent=\(\sqrt{1+x^2}\), hypotenuse=\(\sqrt{x^2+2}\).</p><p><strong>Step 4:</strong> <span class="math-block">\[\cos\alpha=\frac{\sqrt{1+x^2}}{\sqrt{x^2+2}}=\sqrt{\frac{x^2+1}{x^2+2}}\]</p><p><strong>Answer: (A)</strong></p><div class="trap-box"><strong>Trap:</strong> Misidentifying which side is opposite/adjacent in the second triangle.<div class="key-concept"><strong>Key Concept:</strong> Right-triangle method for nested ITF expressions