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
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