Trigonometry & Inverse Trigonometry
Inverse Trigonometric Functions
Grade 12
Question:
<p>$\tan^{-1}(\tan \theta) = \theta$, for all $\theta$ belonging to</p>
<p>(a) $[0, \pi]$</p>
<p>(b) $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] - \{0\}$</p>
<p>(c) $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$</p>
<p>(d) None of these</p>
Step-by-Step Solution
Key Concept: The inverse tangent function has principal branch as an open interval $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ excluding the endpoints.
<p>The principal value branch of $\tan^{-1}$ is $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ (open interval). Therefore, $\tan^{-1}(\tan \theta) = \theta$ holds for all $\theta$ in this interval.</p>
Correct Answer: C