Given: x+y=90°...(1)
( sinx+ siny)2−2 sinx siny=\sin2x+\sin2y+2 sinx siny−2 sinx siny
= \sin2x+\sin2y...(2)
Recall: sinx=\cos(90−x)...(a)
From (1), y=90−x...(b)
Putting (a) and (b) in (2), we have
\sin2x+\sin2y≡\cos2(90−x)+\sin2(90−x)
= 1