Ii
MATU • Trigonometria
MATU_TRI_352
Propio
Enunciado
Paso 1:
Si $\theta = \frac{23\pi}{6}$, entonces halle el valor de $\sec \theta - \tan \theta$.
Si $\theta = \frac{23\pi}{6}$, entonces halle el valor de $\sec \theta - \tan \theta$.
Solución Paso a Paso
1. Convertir o reducir el ángulo:
$$ \theta = \frac{23\pi}{6} = \frac{(24 - 1)\pi}{6} = 4\pi - \frac{\pi}{6} $$
Como $4\pi$ representa dos vueltas completas, el ángulo es equivalente a $-\frac{\pi}{6}$ o $330^\circ$.
2. Calcular las razones:
$$ \begin{aligned} \sec\left(\frac{23\pi}{6}\right) &= \sec\left(-\frac{\pi}{6}\right) = \sec\left(\frac{\pi}{6}\right) = \frac{2}{\sqrt{3}} = \frac{2\sqrt{3}}{3} \\ \tan\left(\frac{23\pi}{6}\right) &= \tan\left(-\frac{\pi}{6}\right) = -\tan\left(\frac{\pi}{6}\right) = -\frac{1}{\sqrt{3}} = -\frac{\sqrt{3}}{3} \end{aligned} $$
3. Operación final:
$$ \sec \theta - \tan \theta = \frac{2\sqrt{3}}{3} - \left(-\frac{\sqrt{3}}{3}\right) = \frac{2\sqrt{3} + \sqrt{3}}{3} = \frac{3\sqrt{3}}{3} = \sqrt{3} $$
$$ \boxed{\sqrt{3}} $$
$$ \theta = \frac{23\pi}{6} = \frac{(24 - 1)\pi}{6} = 4\pi - \frac{\pi}{6} $$
Como $4\pi$ representa dos vueltas completas, el ángulo es equivalente a $-\frac{\pi}{6}$ o $330^\circ$.
2. Calcular las razones:
$$ \begin{aligned} \sec\left(\frac{23\pi}{6}\right) &= \sec\left(-\frac{\pi}{6}\right) = \sec\left(\frac{\pi}{6}\right) = \frac{2}{\sqrt{3}} = \frac{2\sqrt{3}}{3} \\ \tan\left(\frac{23\pi}{6}\right) &= \tan\left(-\frac{\pi}{6}\right) = -\tan\left(\frac{\pi}{6}\right) = -\frac{1}{\sqrt{3}} = -\frac{\sqrt{3}}{3} \end{aligned} $$
3. Operación final:
$$ \sec \theta - \tan \theta = \frac{2\sqrt{3}}{3} - \left(-\frac{\sqrt{3}}{3}\right) = \frac{2\sqrt{3} + \sqrt{3}}{3} = \frac{3\sqrt{3}}{3} = \sqrt{3} $$
$$ \boxed{\sqrt{3}} $$