UPSC NDA I 2018 Mathematics Solved Paper

Concept:The principal value of $\sin^{-1}(\sin\theta)$ must lie in the interval $[-\frac{\pi}{2}, \frac{\pi}{2}]$.If $\theta$ is outside this range, use trigonometric identities to express $\sin\theta$ in terms of an angle within the principal range.Explanation:We need to evaluate $\sin^{-1}(\sin\frac{2\pi}{3})$.First, note that $\frac{2\pi}{3}$ is not in the principal range $[-\frac{\pi}{2}, \frac{\pi}{2}]$ for $\sin^{-1}$.Apply the identity $\sin(\pi - \theta) = \sin \theta$.Since $\frac{2\pi}{3} = \pi - \frac{\pi}{3}$, we have $\sin(\frac{2\pi}{3}) = \sin(\pi - \frac{\pi}{3}) = \sin(\frac{\pi}{3})$.Now, $\frac{\pi}{3}$ lies within $[-\frac{\pi}{2}, \frac{\pi}{2}]$.For any angle $\alpha$ in this interval, $\sin^{-1}(\sin \alpha) = \alpha$.Thus, $\sin^{-1}(\sin\frac{\pi}{3}) = \frac{\pi}{3}$.Therefore, the principal value is $\frac{\pi}{3}$.Answer:$\frac{\pi}{3}$ (Option C).