Step-by-Step Solution
Key Concept: When the angle is outside the principal branch of $\tan^{-1}$, we must reduce it using the periodicity of tangent to find the equivalent angle within $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$.
<p><strong>Step 1:</strong> Note that $2$ radians is approximately $114.6°$, which lies in the second quadrant (between $\frac{\pi}{2}$ and $\pi$).</p><p><strong>Step 2:</strong> Since $\tan$ has period $\pi$, we have $\tan(2) = \tan(2 - \pi)$ where $2 - \pi \approx -1.14$.</p><p><strong>Step 3:</strong> Thus $\tan^{-1}(\tan 2) = 2 - \pi$ (the principal value in $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$).</p><p><strong>Step 4:</strong> Therefore, $\cos\{\tan^{-1}(\tan 2)\} = \cos(2-\pi) = -\cos(2)$.</p>
Correct Answer: D