Given: \(x + y = 90^\circ ... (1)\) \((\sin x + \sin y)^{2} - 2\sin x \sin y = \sin^{2} x + \sin^{2} y + 2\sin x \sin y - 2\sin x \sin y\) = \(\sin^{2} x + \sin^{2} y ... (2)\) Recall: \(\sin x = \cos (90 - x) ... (a)\) From (1), \(y = 90 - x ... (b)\) Putting (a) and (b) in (2), we have \(\sin^{2} x + \sin^{2} y \equiv \cos^{2} (90 - x) + \sin^{2} (90 - x)\) = 1