MHTCET 19 Apr 2026 Shift 2 Paper

Concept:Inverse trigonometric functions return values only within their principal ranges. So we first convert each angle into its equivalent angle lying in the principal range, then evaluate the sum.Explanation:For $\sin^{-1}x$, the principal range is $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$.Since $\sin\frac{5\pi}{6} = \sin\frac{\pi}{6}$, we get:$\sin^{-1}\left(\sin\frac{5\pi}{6}\right) = \frac{\pi}{6}$.For $\cos^{-1}x$, the principal range is $[0, \pi]$.Since $\cos\frac{7\pi}{6} = \cos\frac{5\pi}{6}$, we get:$\cos^{-1}\left(\cos\frac{7\pi}{6}\right) = \frac{5\pi}{6}$.For $\tan^{-1}x$, the principal range is $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$.Since $\tan\frac{2\pi}{3} = \tan\left(-\frac{\pi}{3}\right)$, we get:$\tan^{-1}\left(\tan\frac{2\pi}{3}\right) = -\frac{\pi}{3}$.Now add all three terms:$\frac{\pi}{6} + \frac{5\pi}{6} - \frac{\pi}{3} = \pi - \frac{\pi}{3} = \frac{2\pi}{3}$.Answer:$\frac{2\pi}{3}$, which is Option A.