Inverse Trigonometric Functions

The deciding concept is the principal-range restriction of arcsine, since ${\sin }^{-1}$ is defined to return values only in the closed interval $\left[-\frac{\pi }{2},\frac{\pi }{2}\right]$, a restriction made necessary because sine is many-to-one and would otherwise have no single inverse. The angle $\frac{2\pi }{3}$ lies outside this principal interval, so ${\sin }^{-1}\left(\sin \frac{2\pi }{3}\right)$ cannot simply return $\frac{2\pi }{3}$, which immediately marks the assertion as false. To find the correct value, use the supplementary-angle relation $\sin \frac{2\pi }{3}=\sin \left(\pi -\frac{2\pi }{3}\right)=\sin \frac{\pi }{3}=\frac{\sqrt{3}}{2}$; because $\frac{\pi }{3}$ does lie inside the principal range, ${\sin }^{-1}\frac{\sqrt{3}}{2}=\frac{\pi }{3}$. The reason is likewise false, because arcsine inverts sine only on its restricted domain, not everywhere on the real line. The option claiming both statements true ignores the range entirely. The assertion-true option is impossible given the range bound. The assertion-false-but-reason-true option wrongly endorses a reason that is itself incorrect. This is the canonical inverse-composition pitfall repeatedly tested in JEE Advanced. As a final consistency check, $\sin \frac{\pi }{3}=\sin \frac{2\pi }{3}=\frac{\sqrt{3}}{2}$ and $\frac{\pi }{3}$ alone lies within $\left[-\frac{\pi }{2},\frac{\pi }{2}\right]$, so it is indeed the unique principal output, confirming both that the assertion is false and that the corrected value is $\frac{\pi }{3}$.