If $x \in \mathbb{R}$, then $f(x) = \cos^{-1}\left(\frac{1 - x^2}{1 + x^2}\right)$ is equal to
- A. $2\tan^{-1}x$
- B. $\begin{cases} 2\tan^{-1}x, & x \ge 0 \\ -2\tan^{-1}x, & x \le 0 \end{cases}$
- C. $\begin{cases} \pi + 2\tan^{-1}x, & x \ge 0 \\ -\pi + 2\tan^{-1}x, & x \le 0 \end{cases}$
- D. None of these
✅ Correct Answer: $\begin{cases} 2\tan^{-1}x, & x \ge 0 \\ -2\tan^{-1}x, & x \le 0 \end{cases}$
Explanation
Using $\tan(\theta) = x \Rightarrow \cos(\theta) = \frac{1 - x^2}{1 + x^2}$. Hence $\cos^{-1}\left(\frac{1 - x^2}{1 + x^2}\right) = 2|\tan^{-1}x|$.
🎯 Happy Preparation — ACME Academy
