1
0
2
-1
Correct answer is A
Given: \(x + y = 90° ... (1)\)
\((\sin x + \sin y)^{2} - 2\sin x \sin y = \sin^{2} x + \sin^{2} y + 2\sin x \sin y - 2\sin x \sin y\)
= \(\sin^{2} x + \sin^{2} y ... (2)\)
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