$\cos^2(x - y) = \sin(2x)\sin(2y) = (\cos x \cos y + \sin x \sin y)^2$ $ = 2 \sin x \cos x \cdot 2 \sin y \cos y$ $= ( \cos x \cos y)^2 + 2 \sin x \cos x \cdot \sin y \cos y+ (\sin x \sin y)^2 = 4 \sin x \cos x \sin y \cos y$ $\implies ( \cos x \cos y)^2 - 2 \sin x \cos x \cdot \sin y \cos y+ (\sin x \sin y)^2 = 0$ $ =(\cos x \cos y - \sin x \sin y)^2 = 0 \implies \cos x \cos y - \sin x \sin y = 0$ $= \cos(x + y) = 0 = \cos(x + y) = \cos(\frac{\pi}{2})$ $\implies \boxed{x + y = \dfrac{\pi}{2}}$ $\textbf{Resposta : Alternativa A}$

Em ambos os lados vamos multiplicar por 2 e em seguida subtrair 1: $$2cos²(x-y)-1=2sen2x.sen2y-1$$ $$cos(2x-2y)=2sen2x.sen2y-1$$ $$cos2x.cos2y+sen2x.sen2y=2.sen2x.sen2y-1$$ $$cos2x.cos2y-sen2x.sen2y=-1$$ $$cos(2x+2y)=-1$$ Assim: $$2x+2y=\pi + 2k\pi$$ $$x+y=\frac{\pi}{2}+k\pi$$ $$x+y=\frac{\pi}{2}$$