Given that the inverse trigonometric function assumes principal values only. Let \(x\), \(y\) be any two real numbers in…

Let \({\cos }^{-1}x=A\) and \({\sin }^{-1}y=B\). Then \(x=\cos A\) and \(y=\sin B\).
given condition is \(A-B=\alpha\).
The expression is \(E={x}^{2}+{y}^{2}+2xy\sin \alpha\).
Substitute \(x,y\), and \(\alpha\) :

\(E={\cos }^{2}A+{\sin }^{2}B+2\cos A\sin B\sin (A-B)\)

Using the identity \(\sin (A-B)=\sin A\cos B-\cos A\sin B\) :

\(E={\cos }^{2}A+{\sin }^{2}B+2\cos A\sin B(\sin A\cos B-\cos A\sin B)\)

\(E={\cos }^{2}A+{\sin }^{2}B+2\sin A\cos A\sin B\cos B-2{\cos }^{2}A{\sin }^{2}B\)

\(E={\cos }^{2}A\left(1-{\sin }^{2}B\right)+{\sin }^{2}B\left(1-{\cos }^{2}A\right)+2\sin A\cos A\sin B\cos B\)

\(E={\cos }^{2}A{\cos }^{2}B+{\sin }^{2}A{\sin }^{2}B+2\sin A\sin B\cos A\cos B\\ E=(\cos A\cos B+\sin A\sin B{)}^{2}=(\cos (A-B){)}^{2}={\cos }^{2}\alpha\)

The minimum value of \({\cos }^{2}\alpha \text{ is }0\)

At \(\alpha=\frac{\pi}{2}\)