MHTCET 13 Apr 2026 Shift 2 Paper

Concept:Use the principal values of standard inverse trigonometric functions and simplify the resulting fraction.Explanation:Evaluate the numerator first.$\tan^{-1}(1) = \frac{\pi}{4}$, since $\tan\frac{\pi}{4} = 1$.$\cos^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{3}$, since $\cos\frac{\pi}{3} = \frac{1}{2}$.$\sin^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{6}$, since $\sin\frac{\pi}{6} = \frac{1}{2}$.So the numerator is $\frac{\pi}{4} + \frac{\pi}{3} + \frac{\pi}{6}$.Now, $\frac{\pi}{3} + \frac{\pi}{6} = \frac{2\pi}{6} + \frac{\pi}{6} = \frac{3\pi}{6} = \frac{\pi}{2}$.Hence, numerator $= \frac{\pi}{4} + \frac{\pi}{2} = \frac{\pi}{4} + \frac{2\pi}{4} = \frac{3\pi}{4}$.Next, evaluate the denominator.$\tan^{-1}(\sqrt{3}) = \frac{\pi}{3}$, since $\tan\frac{\pi}{3} = \sqrt{3}$.Also, $\sec^{-1}(-2) = \cos^{-1}\left(-\frac{1}{2}\right)$.Since $\cos\frac{2\pi}{3} = -\frac{1}{2}$ and $\frac{2\pi}{3} \in [0, \pi]$, we get $\sec^{-1}(-2) = \frac{2\pi}{3}$.Thus, denominator $= \frac{\pi}{3} - \frac{2\pi}{3} = -\frac{\pi}{3}$.Now compute the required value.$$\frac{\frac{3\pi}{4}}{-\frac{\pi}{3}} = \frac{3\pi}{4} \times \frac{-3}{\pi} = -\frac{9}{4}$$Answer:$-\frac{9}{4}$Therefore, the correct option is D.