Question:

<p><strong>Step 1: Identify the constraint.</strong> For $\cos^{-1}\left(\frac{1-x^2}{1+x^2}\right)$ to be defined, we need:<br>$$-1 \leq \frac{1-x^2}{1+x^2} \leq 1$$</p><p><strong>Step 2: Use substitution $x = \tan\theta$.</strong> When $x = \tan\theta$, we have:<br>$$\frac{1-x^2}{1+x^2} = \frac{1-\tan^2\theta}{1+\tan^2\theta} = \frac{\cos^2\theta - \sin^2\theta}{\cos^2\theta + \sin^2\theta} = \cos(2\theta)$$</p><p><strong>Step 3: Analyze the right inequality.</strong> We need $\frac{1-x^2}{1+x^2} \leq 1$:<br>$$1-x^2 \leq 1+x^2$$<br>$$-x^2 \leq x^2$$<br>$$-2x^2 \leq 0$$<br>This is always true for all $x \in \mathbb{R}$.</p><p><strong>Step 4: Analyze the left inequality.</strong> We need $\frac{1-x^2}{1+x^2} \geq -1$:<br>$$1-x^2 \geq -(1+x^2)$$<br>$$1-x^2 \geq -1-x^2$$<br>$$1 \geq -1$$<br>This is always true for all $x \in \mathbb{R}$.</p><p><strong>Step 5: Verify with the cosine substitution.</strong> Since $\cos(2\theta) \in [-1, 1]$ for all $\theta$, and $\frac{1-x^2}{1+x^2} = \cos(2\theta)$ where $x = \tan\theta$, the expression is valid for all real $x$ where $\tan\theta$ is defined. Note that $1+x^2 > 0$ for all real $x$, so the denominator is never zero.</p><p><strong>Step 6: Conclusion.</strong> The expression $\frac{1-x^2}{1+x^2}$ is always in $[-1, 1]$ for all $x \in \mathbb{R}$, so the domain is $\mathbb{R}$.</p><p><strong>∴ Answer:</strong> A</p>