Carrillo Campos
01/14/2023 · Primary School
Temo: I dentidades trigonemé Resolver las siguientes I dentidades 1. \( \cot \alpha(\tan \alpha+\cot \alpha)=\csc ^{2} \alpha \)
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Step-by-step Solution
Para resolver la identidad \( \cot \alpha(\tan \alpha+\cot \alpha)=\csc ^{2} \alpha \), comenzamos descomponiendo los términos en el lado izquierdo.
Recordemos que:
- \( \cot \alpha = \frac{\cos \alpha}{\sin \alpha} \)
- \( \tan \alpha = \frac{\sin \alpha}{\cos \alpha} \)
- \( \csc \alpha = \frac{1}{\sin \alpha} \), por lo que \( \csc^2 \alpha = \frac{1}{\sin^2 \alpha} \)
Ahora, sustituimos \( \tan \alpha \) y \( \cot \alpha \) en la expresión:
\[
\cot \alpha(\tan \alpha + \cot \alpha) = \cot \alpha\left(\frac{\sin \alpha}{\cos \alpha} + \frac{\cos \alpha}{\sin \alpha}\right)
\]
Simplificamos el término dentro del paréntesis:
\[
\frac{\sin \alpha}{\cos \alpha} + \frac{\cos \alpha}{\sin \alpha} = \frac{\sin^2 \alpha + \cos^2 \alpha}{\sin \alpha \cos \alpha}
\]
Dado que \( \sin^2 \alpha + \cos^2 \alpha = 1 \), podemos simplificar aún más:
\[
\frac{1}{\sin \alpha \cos \alpha}
\]
Ahora, sustituimos esto de nuevo en la expresión original:
\[
\cot \alpha \cdot \frac{1}{\sin \alpha \cos \alpha} = \frac{\cos \alpha}{\sin \alpha} \cdot \frac{1}{\sin \alpha \cos \alpha} = \frac{1}{\sin^2 \alpha}
\]
Esto es igual a \( \csc^2 \alpha \):
\[
\frac{1}{\sin^2 \alpha} = \csc^2 \alpha
\]
Por lo tanto, hemos demostrado que:
\[
\cot \alpha(\tan \alpha + \cot \alpha) = \csc^2 \alpha
\]
La identidad es verdadera.
Quick Answer
La identidad \( \cot \alpha(\tan \alpha+\cot \alpha)=\csc ^{2} \alpha \) se demuestra correctamente utilizando las definiciones de \( \cot \alpha \), \( \tan \alpha \), y \( \csc \alpha \).
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