Question:

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