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MATU • Trigonometria
MATU_TRI_334
Guía de ejercicios
Enunciado
Paso 1:
Si $\tan^2 \theta = 1 - e^2$, demuestre que $\sec \theta + \tan^3 \theta \cdot \csc \theta = (2 - e^2)^{3/2}$.
Si $\tan^2 \theta = 1 - e^2$, demuestre que $\sec \theta + \tan^3 \theta \cdot \csc \theta = (2 - e^2)^{3/2}$.
Solución Paso a Paso
1. Simplificación de la expresión:
$E = \sec \theta + \tan^3 \theta \cdot \csc \theta$
$$ E = \sec \theta + \frac{\sin^3 \theta}{\cos^3 \theta} \cdot \frac{1}{\sin \theta} = \sec \theta + \frac{\sin^2 \theta}{\cos^3 \theta} $$
Factorizamos $\sec \theta$ (o $1/\cos \theta$):
$$ E = \frac{1}{\cos \theta} \left( 1 + \frac{\sin^2 \theta}{\cos^2 \theta} \right) = \sec \theta (1 + \tan^2 \theta) $$
Como $1 + \tan^2 \theta = \sec^2 \theta$:
$$ E = \sec \theta \cdot \sec^2 \theta = \sec^3 \theta = (\sec^2 \theta)^{3/2} $$
2. Sustitución de datos:
Sabemos que $\sec^2 \theta = 1 + \tan^2 \theta$. Sustituimos $\tan^2 \theta = 1 - e^2$:
$$ \sec^2 \theta = 1 + (1 - e^2) = 2 - e^2 $$
Sustituyendo en $E$:
$$ E = (2 - e^2)^{3/2} $$
3. Resultado:
$$ \boxed{\sec \theta + \tan^3 \theta \cdot \csc \theta = (2 - e^2)^{3/2}} $$
$E = \sec \theta + \tan^3 \theta \cdot \csc \theta$
$$ E = \sec \theta + \frac{\sin^3 \theta}{\cos^3 \theta} \cdot \frac{1}{\sin \theta} = \sec \theta + \frac{\sin^2 \theta}{\cos^3 \theta} $$
Factorizamos $\sec \theta$ (o $1/\cos \theta$):
$$ E = \frac{1}{\cos \theta} \left( 1 + \frac{\sin^2 \theta}{\cos^2 \theta} \right) = \sec \theta (1 + \tan^2 \theta) $$
Como $1 + \tan^2 \theta = \sec^2 \theta$:
$$ E = \sec \theta \cdot \sec^2 \theta = \sec^3 \theta = (\sec^2 \theta)^{3/2} $$
2. Sustitución de datos:
Sabemos que $\sec^2 \theta = 1 + \tan^2 \theta$. Sustituimos $\tan^2 \theta = 1 - e^2$:
$$ \sec^2 \theta = 1 + (1 - e^2) = 2 - e^2 $$
Sustituyendo en $E$:
$$ E = (2 - e^2)^{3/2} $$
3. Resultado:
$$ \boxed{\sec \theta + \tan^3 \theta \cdot \csc \theta = (2 - e^2)^{3/2}} $$