MHTCET 15 Apr 2026 Shift 1 Paper

Concept:The inverse cosine function returns values only in $[0, \pi]$, so we must rewrite $\sin x$ as a cosine of an angle that lies in this principal range.Explanation:Given $y = \cos^{-1}(\sin x)$, where $\frac{\pi}{2} < x < \pi$.We know the identity $\sin x = \cos\left(\frac{\pi}{2} - x\right)$.But since $x > \frac{\pi}{2}$, we have $\frac{\pi}{2} - x < 0$.Using $\cos(-\theta) = \cos \theta$, we rewrite it as:$\sin x = \cos\left(x - \frac{\pi}{2}\right)$.Now, for $\frac{\pi}{2} < x < \pi$, it follows that $0 < x - \frac{\pi}{2} < \frac{\pi}{2}$.This angle $x - \frac{\pi}{2}$ lies within the principal value range $[0, \pi]$ of $\cos^{-1}$.Hence, $y = \cos^{-1}\left[\cos\left(x - \frac{\pi}{2}\right)\right] = x - \frac{\pi}{2}$.Differentiating both sides with respect to $x$, we get:$\frac{dy}{dx} = 1$.So the correct option is C.Answer:$\frac{dy}{dx} = \boxed{1}$Option C.