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\( \begin{array}{l}\text { Reduce: } \\ M=\frac{(a+b)^{2} \cdot \cos 360^{\circ}+(a-b)^{2} \operatorname{sen} 270^{\circ}}{a \operatorname{sen} 180^{\circ}+a b \operatorname{sen} 270^{\circ}+b \operatorname{sen} 360^{\circ}} \\ \begin{array}{llll}\text { A) } 1 & \text { B) } 4 & \text { C) } 2 & \text { D) }-4\end{array} \\ \text { E) } 3\end{array} l \).

Ask by Frazier Mann. in Peru
Mar 15,2025

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La respuesta es -4, opción D.

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To simplify the expression for \( M \): 1. We first compute the trigonometric values: - \( \cos 360^\circ = 1 \) - \( \sin 270^\circ = -1 \) - \( \sin 180^\circ = 0 \) - \( \sin 360^\circ = 0 \) 2. Plugging these values into the equation: \[ M = \frac{(a+b)^{2} \cdot 1 + (a-b)^{2} \cdot (-1)}{a \cdot 0 + a b \cdot (-1) + b \cdot 0} \] This simplifies to: \[ M = \frac{(a+b)^{2} - (a-b)^{2}}{-ab} \] 3. Now, use the difference of squares for the numerator: \[ M = \frac{[(a+b) + (a-b)][(a+b) - (a-b)]}{-ab} = \frac{(2a)(2b)}{-ab} \] 4. Simplifying further yields: \[ M = \frac{4ab}{-ab} = -4 \] Thus, the answer is \( \boxed{-4} \).

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