JEE Main 2024 — Inverse Trigonometric Functions Question with Solution

$\begin{aligned} & \cos ^{-1} x-\left(\frac{\pi}{2}-\cos ^{-1} y\right)=\alpha \\ & \cos ^{-1} x+\cos ^{-1} y=\frac{\pi}{2}+\alpha \\ & \alpha \in\left[-\frac{\pi}{2}, \pi\right], \frac{\pi}{2}+\alpha \in\left[0, \frac{3 \pi}{2}\right] \\ & \cos ^{-1}\left(x y-\sqrt{1-x^2} \sqrt{1-y^2}\right)=\frac{\pi}{2}+\alpha \\ & x y-\sqrt{1-x^2} \sqrt{1-y^2}=-\sin \alpha \\ & (x y+\sin \alpha)=\left(1-x^2\right)\left(1-y^2\right) \\ & x^2 y^2+2 x y \sin a+\sin ^2 a=1-x^2-y^2+x^2 y^2 \\ & x^2+y^2+2 x y \sin \alpha=1-\sin ^2 \alpha \\ & x^2+y^2+2 x y \sin \alpha=\cos ^2 \alpha \end{aligned}$<br/>Min.valueof$\cos ^2 \alpha=0$ At $\alpha =\frac{\pi }{2}$