JAMB Mathematics 2015 Paper

Given: $x + y = 90^{\circ} \dots (1)$

$(\sin x + \sin y)^2 - 2\sin x \sin y = \sin^2 x + \sin^2 y + 2\sin x \sin y - 2\sin x \sin y$

= $\sin^2 x + \sin^2 y \dots (2)$

Recall: $\sin x = \cos(90 - x) \dots (a)$

From (1), $y = 90 - x \dots (b)$

Putting (a) and (b) in (2), we have

$\sin^2 x + \sin^2 y \equiv \cos^2 (90 - x) + \sin^2 (90 - x)$

= 1