Deleon Nichols
03/05/2023 · High School

2) Halla el resultado de la siguiente expresión: \( B=\left(\cos 270^{\circ}\right)^{\operatorname{sen} 90^{\circ}}-\frac{\tan 360^{\circ}}{\cos 0^{\circ}}+5\left(\sec 180^{\circ}\right)^{\cot 270^{\circ}} \)

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Para resolver la expresión \( B=\left(\cos 270^{\circ}\right)^{\operatorname{sen} 90^{\circ}}-\frac{\tan 360^{\circ}}{\cos 0^{\circ}}+5\left(\sec 180^{\circ}\right)^{\cot 270^{\circ}} \), primero calculamos cada una de las funciones trigonométricas involucradas: 1. \( \cos 270^{\circ} = 0 \) 2. \( \sin 90^{\circ} = 1 \) 3. \( \tan 360^{\circ} = 0 \) 4. \( \cos 0^{\circ} = 1 \) 5. \( \sec 180^{\circ} = -1 \) 6. \( \cot 270^{\circ} = 0 \) Ahora sustituimos estos valores en la expresión original: \[ B = (0)^1 - \frac{0}{1} + 5(-1)^0 \] \[ B = 0 - 0 + 5(1) \] \[ B = 0 + 5 \] \[ B = 5 \] Por lo tanto, el resultado de la expresión es \( B = 5 \).

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El resultado de la expresión es \( B = 5 \).
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