Kerala KEAM 5 June 2024 Solved Paper

We are asked to find the value of $\sin^{-1}\left(\sin \left(\frac{5\pi}{6}\right)\right)$.First, recall that the function $\sin^{-1}(x)$ (inverse sine) is defined on the principal range $\left[ -\frac{\pi}{2}, \frac{\pi}{2} \right]$. This means that for any angle $\theta$, $\sin^{-1}(\sin \theta)$ will return the value of $\theta$ within this range.Now, consider the angle $\frac{5\pi}{6}$. This angle is greater than $\frac{\pi}{2}$, so we need to adjust it to fall within the principal range. Since $\sin \left( \frac{5\pi}{6} \right) = \sin \left( \frac{\pi}{6} \right)$, we have:$$\sin^{-1} \left( \sin \left( \frac{5\pi}{6} \right) \right) = \sin^{-1} \left( \sin \left( \frac{\pi}{6} \right) \right)$$Therefore, the result is $\frac{\pi}{6}$, as $\frac{\pi}{6}$ is within the principal range of $\sin^{-1}$.Thus, the correct answer is option (E), $\frac{\pi}{6}$. Quick Tip: For inverse trigonometric functions, always ensure the result is within the range of the function. If the argument is outside the range, find the corresponding equivalent angle within the domain.