MHTCET 18 Apr 2026 Shift 1 Paper

Concept: Simplify the inverse trigonometric expressions using standard identities, then apply the chain rule to find $\frac{dy}{dx}$.Explanation:Since $t \in (0,1)$, we have $\frac{1+t^2}{2t} > 1$. Given $x = \csc^{-1}\left(\frac{1+t^2}{2t}\right)$, we get $\sin x = \frac{2t}{1+t^2}$. Using $\sin(2\tan^{-1}t) = \frac{2t}{1+t^2}$ and the given range of $t$, we obtain $x = 2\tan^{-1}t$. Thus, $\frac{dx}{dt} = \frac{2}{1+t^2}$. For $y = \cot^{-1}\left(\frac{\sqrt{1-t^2}}{t}\right)$, let $\theta = \sin^{-1}t$. Then $\sin\theta = t$ and $\cos\theta = \sqrt{1-t^2}$, so $\frac{\sqrt{1-t^2}}{t} = \cot\theta$. Therefore, $y = \sin^{-1}t$, and hence $\frac{dy}{dt} = \frac{1}{\sqrt{1-t^2}}$. By the chain rule, $$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{\frac{1}{\sqrt{1-t^2}}}{\frac{2}{1+t^2}} = \frac{1+t^2}{2\sqrt{1-t^2}}.$$ So $f(t) = \frac{1+t^2}{2\sqrt{1-t^2}}$. Substituting $t = \tan\alpha$: $$f(\tan\alpha) = \frac{1+\tan^2\alpha}{2\sqrt{1-\tan^2\alpha}} = \frac{\sec^2\alpha}{2\sqrt{\frac{\cos 2\alpha}{\cos^2\alpha}}} = \frac{\sec\alpha}{2\sqrt{\cos 2\alpha}}.$$ This simplification is valid since $\alpha \in \left(0, \frac{\pi}{4}\right)$, giving $\cos\alpha > 0$ and $\cos 2\alpha > 0$.Answer:The correct option is B: $\frac{\sec \alpha}{2 \sqrt{\cos 2 \alpha}}$.