Page Wyatt
09/03/2023 · Elementary School
\( \operatorname { Sec } ^ { 2 } \alpha ( 1 - \cos ^ { 2 } \alpha ) = \operatorname { Tg } ^ { 2 } \alpha \)
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Para demostrar la identidad \( \sec^2 \alpha (1 - \cos^2 \alpha) = \tan^2 \alpha \), comenzamos recordando algunas identidades trigonométricas fundamentales.
Sabemos que:
1. \( \sec^2 \alpha = \frac{1}{\cos^2 \alpha} \)
2. \( \tan^2 \alpha = \frac{\sin^2 \alpha}{\cos^2 \alpha} \)
3. La identidad pitagórica: \( \sin^2 \alpha + \cos^2 \alpha = 1 \), lo que implica que \( \sin^2 \alpha = 1 - \cos^2 \alpha \).
Ahora, sustituyamos \( 1 - \cos^2 \alpha \) en la expresión original:
\[
1 - \cos^2 \alpha = \sin^2 \alpha
\]
Sustituyendo esto en la ecuación:
\[
\sec^2 \alpha (1 - \cos^2 \alpha) = \sec^2 \alpha \sin^2 \alpha
\]
Ahora, sustituimos \( \sec^2 \alpha \):
\[
\sec^2 \alpha \sin^2 \alpha = \frac{1}{\cos^2 \alpha} \sin^2 \alpha
\]
Esto se puede simplificar a:
\[
\frac{\sin^2 \alpha}{\cos^2 \alpha} = \tan^2 \alpha
\]
Por lo tanto, hemos demostrado que:
\[
\sec^2 \alpha (1 - \cos^2 \alpha) = \tan^2 \alpha
\]
Así que la identidad es verdadera.
Quick Answer
Demostramos que \( \sec^2 \alpha (1 - \cos^2 \alpha) = \tan^2 \alpha \) usando identidades trigonométricas básicas.
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