MHTCET 16 Apr 2026 Shift 2 Paper

Concept:Simplify the trigonometric expression using half-angle identities, then use the principal range of $\cot^{-1}$ to find the angle.Explanation:For $x \in \left(0, \frac{\pi}{2}\right)$, we have $\frac{x}{2} \in \left(0, \frac{\pi}{4}\right)$.In this interval, $\cos \frac{x}{2} > \sin \frac{x}{2} > 0$.Use the standard identities:$$1+\sin x = \left(\cos \frac{x}{2} + \sin \frac{x}{2}\right)^2$$$$1-\sin x = \left(\cos \frac{x}{2} - \sin \frac{x}{2}\right)^2$$So we get:$$\sqrt{1+\sin x} = \cos \frac{x}{2} + \sin \frac{x}{2}$$$$\sqrt{1-\sin x} = \cos \frac{x}{2} - \sin \frac{x}{2}$$Substitute these into the given fraction:$$\frac{\sqrt{1-\sin x}+\sqrt{1+\sin x}}{\sqrt{1-\sin x}-\sqrt{1+\sin x}} = \frac{2\cos \frac{x}{2}}{-2\sin \frac{x}{2}} = -\cot \frac{x}{2}$$Thus the expression becomes:$$\cot^{-1}\left(-\cot \frac{x}{2}\right)$$Now, $\cot\left(\pi - \frac{x}{2}\right) = -\cot \frac{x}{2}$.Also, $\pi - \frac{x}{2}$ lies in $\left(\frac{\pi}{2}, \pi\right)$, which is within the principal value range $(0, \pi)$ of $\cot^{-1}$.Therefore:$$\cot^{-1}\left(-\cot \frac{x}{2}\right) = \pi - \frac{x}{2}$$Answer:$\pi - \frac{x}{2}$, which is Option D.