MHTCET 10 May 2024 Shift 1 Paper

Concept:The inverse tangent function, $\tan^{-1}x$, returns values only in the principal branch $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$.So, when the angle inside $\tan$ lies outside this range, we must first reduce it to an equivalent angle within the principal range.Explanation:We need to evaluate $\tan^{-1}\left(\tan \frac{7\pi}{6}\right)$.Write $\frac{7\pi}{6}$ as $\pi + \frac{\pi}{6}$.Using the identity $\tan(\pi + \theta) = \tan\theta$, we get:$\tan\left(\frac{7\pi}{6}\right) = \tan\left(\pi + \frac{\pi}{6}\right) = \tan\frac{\pi}{6}$.Now, $\frac{\pi}{6}$ lies in the principal range of $\tan^{-1}$, which is $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$.Therefore, $\tan^{-1}\left(\tan\frac{\pi}{6}\right) = \frac{\pi}{6}$.No further adjustment is needed because the angle $\frac{\pi}{6}$ already lies in the correct interval.Answer:$\tan^{-1}\left(\tan \frac{7\pi}{6}\right) = \frac{\pi}{6}$Correct option: B. $\frac{\pi}{6}$