1. The problem states: Given that $x + y = 90^\circ$, find the value of the expression $$\frac{\cos(2x + y) + \sin(3x + 2y)}{\cos y}$$ 2. First, use the given relation $x + y = 90^\circ$ to express angles in terms of one variable. Since $y = 90^\circ - x$, substitute this into the expression. 3. Substitute $y = 90^\circ - x$: $$\cos(2x + y) = \cos(2x + 90^\circ - x) = \cos(x + 90^\circ)$$ $$\sin(3x + 2y) = \sin(3x + 2(90^\circ - x)) = \sin(3x + 180^\circ - 2x) = \sin(x + 180^\circ)$$ 4. Use trigonometric identities: - $\cos(\theta + 90^\circ) = -\sin \theta$ - $\sin(\theta + 180^\circ) = -\sin \theta$ So, $$\cos(x + 90^\circ) = -\sin x$$ $$\sin(x + 180^\circ) = -\sin x$$ 5. Substitute back: $$\frac{\cos(2x + y) + \sin(3x + 2y)}{\cos y} = \frac{-\sin x + (-\sin x)}{\cos y} = \frac{-2 \sin x}{\cos y}$$ 6. Recall $y = 90^\circ - x$, so $\cos y = \cos(90^\circ - x) = \sin x$. 7. Substitute $\cos y = \sin x$: $$\frac{-2 \sin x}{\sin x} = -2$$ 8. Therefore, the value of the expression is $-2$. Final answer: A) $-2$