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