MHTCET 13 Apr 2026 Shift 1 Paper

Concept:Use the principal value ranges of inverse trigonometric functions to evaluate each term.Explanation:First, find $\tan^{-1}(\sqrt{3})$.Since $\tan \frac{\pi}{3} = \sqrt{3}$ and $\frac{\pi}{3}$ lies in the principal range of $\tan^{-1}$, we get $\tan^{-1}(\sqrt{3}) = \frac{\pi}{3}$.Next, evaluate $\sec^{-1}(-2)$.We need an angle $\theta$ such that $\sec \theta = -2$, which means $\cos \theta = -\frac{1}{2}$.In the principal range of $\sec^{-1}$, the angle is $\frac{2\pi}{3}$.So, $\sec^{-1}(-2) = \frac{2\pi}{3}$.Now, $\sin^{-1}\left(-\frac{1}{2}\right) = -\frac{\pi}{6}$, because $\sin\left(-\frac{\pi}{6}\right) = -\frac{1}{2}$.Substitute these values into the expression:$$\frac{\pi}{3} + \frac{2\pi}{3} - \left(-\frac{\pi}{6}\right)$$This simplifies to:$$\pi + \frac{\pi}{6} = \frac{7\pi}{6}$$Answer:The value is $\frac{7\pi}{6}$, which is Option C.