14). - Simplifica : \[ \begin{array}{lll}\operatorname{Sec} \frac{\pi}{3} \cdot \operatorname{Tag} \frac{\pi}{4} \cdot \operatorname{Ctg} \frac{\pi}{6} \\ \operatorname{Tg} \frac{\pi}{3} \cdot \operatorname{Sec} \frac{\pi}{4} \cdot \operatorname{Ctg} \frac{\pi}{4}\end{array} \] \( \begin{array}{lll}\text { A) } 0 & \text { B) } \sqrt{2} & \text { C) } 4 \sqrt{3} \\ \text { D) } 7 \sqrt{6} & \text { E) } \sqrt{6} & \end{array} \)

Para simplificar la expresión dada: \[ \frac{\operatorname{Sec} \frac{\pi}{3} \cdot \operatorname{Tag} \frac{\pi}{4} \cdot \operatorname{Ctg} \frac{\pi}{6}}{\operatorname{Tg} \frac{\pi}{3} \cdot \operatorname{Sec} \frac{\pi}{4} \cdot \operatorname{Ctg} \frac{\pi}{4}} \] Primero, recordemos las definiciones y valores de las funciones trigonométricas involucradas: 1. **Secante**: \(\operatorname{Sec} \theta = \frac{1}{\cos \theta}\) 2. **Tangente**: \(\operatorname{Tag} \theta = \tan \theta = \frac{\sin \theta}{\cos \theta}\) 3. **Cotangente**: \(\operatorname{Ctg} \theta = \cot \theta = \frac{\cos \theta}{\sin \theta}\) Ahora, calculemos cada función trigonométrica: - \(\operatorname{Sec} \frac{\pi}{3} = \frac{1}{\cos \frac{\pi}{3}} = \frac{1}{\frac{1}{2}} = 2\) - \(\operatorname{Tag} \frac{\pi}{4} = \tan \frac{\pi}{4} = 1\) - \(\operatorname{Ctg} \frac{\pi}{6} = \cot \frac{\pi}{6} = \frac{\cos \frac{\pi}{6}}{\sin \frac{\pi}{6}} = \frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}} = \sqrt{3}\) Ahora, para el denominador: - \(\operatorname{Tg} \frac{\pi}{3} = \tan \frac{\pi}{3} = \sqrt{3}\) - \(\operatorname{Sec} \frac{\pi}{4} = \frac{1}{\cos \frac{\pi}{4}} = \frac{1}{\frac{\sqrt{2}}{2}} = \sqrt{2}\) - \(\operatorname{Ctg} \frac{\pi}{4} = \cot \frac{\pi}{4} = 1\) Ahora sustituimos estos valores en la expresión original: \[ \frac{2 \cdot 1 \cdot \sqrt{3}}{\sqrt{3} \cdot \sqrt{2} \cdot 1} \] Simplificamos: \[ = \frac{2\sqrt{3}}{\sqrt{3}\sqrt{2}} = \frac{2\sqrt{3}}{\sqrt{6}} = \frac{2\sqrt{3}}{\sqrt{3 \cdot 2}} = \frac{2\sqrt{3}}{\sqrt{3}\sqrt{2}} = \frac{2}{\sqrt{2}} = 2 \cdot \frac{\sqrt{2}}{2} = \sqrt{2} \] Por lo tanto, la respuesta simplificada es: \[ \sqrt{2} \] La opción correcta es **B) \(\sqrt{2}\)**.