If \(x + y = 90\)
Solve for \((\sin x + \sin y)^2 - 2\sin x \sin y\)
\(x + y = 90\) ---(1)
\((\sin x + \sin y)^2 - 2\sin x \sin y\) ----(2)
\(x = 90 - y\) ---(3)
Expanding eqn (2)
Therefore: \(\sin^{2}x + \sin^{2}y + 2\sin x \sin y - 2\sin x \sin y\)
=> \(\sin^{2}x + \sin^{2}y\) --- (4)
Substitute the value of \(x\) in eqn (3) into eqn (4)
=> \(\sin^{2}(90-y) + \sin^{2}y\) ---(5)
Expanding eqn (5)
=> \(\sin^{2}(90) - \sin^{2}y + \sin^{2}y\)
=> \(\sin^{2}(90) + 0\)
=> \(\sin^{2}(90)\)
=> \((\sin(90))^{2}\)
=> \(1^2\)
= \(1\)

Since x + y = 90, x and y must be real numbers that satisfy the equation. But in cases like this, x and y must be special angles. Examples of special angles are 30°, 45°, 60° and 90° such that their value can be estimated with ease.
x and y can either be 45° and 45° or 30° and 60°.whichever is your choice the answer still awaits you.
Let x=30° and y=60°
Note that sin30°=½ and sin60°=\( \cfrac{\sqrt{3}}{2} \)
\( (\sin x + \sin y)^{2} -2\sin x \sin y \)
\( = (\sin 30° + \sin 60°)^{2} -2(\sin 30°)(\sin 60°) \)
\( =(½ + \cfrac{\sqrt{3}}{2})^{2} - 2(½)( \cfrac{\sqrt{3}}{2}) \)\( (\cfrac{1 + \sqrt{3}}{2})^{2} -\cfrac{2\sqrt{3}}{4} \)\( =(\cfrac{4 + 2\sqrt{3}}{4}) - \cfrac{2\sqrt{3}}{4} \)
\( =\cfrac{4 +2\sqrt{3} - 2\sqrt{3}}{4} =\cfrac{4}{4} =1 \)
Option A