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verificar (1-tan^4(x))/(1-tan^2(x))=sec^2(x)

Solución

verificar

\frac{1 - tan ^{4} ( x )}{1 - tan ^{2} ( x )} = sec^{2} (x)

Solución

Verdadero

Pasos de solución

\frac{1 - tan ^{4} ( x )}{1 - tan ^{2} ( x )} = sec^{2} (x)

Manipular el lado derecho

\frac{1 - tan ^{4} ( x )}{1 - tan ^{2} ( x )}

Simplificar

\frac{1 - tan ^{4} ( x )}{1 - tan ^{2} ( x )} : tan^{2} (x) + 1

\frac{1 - tan ^{4} ( x )}{1 - tan ^{2} ( x )}

Factorizar

1 - tan^{4} (x) : - (tan^{2} (x) + 1) (tan (x) + 1) (tan (x) - 1)

1 - tan^{4} (x)

Factorizar el termino común

- 1

= - (tan^{4} (x) - 1)

Factorizar

tan^{4} (x) - 1 : (tan^{2} (x) + 1) (tan (x) + 1) (tan (x) - 1)

tan^{4} (x) - 1

Reescribir

tan^{4} (x) - 1

como

(tan^{2} (x))^{2} - 1^{2}

tan^{4} (x) - 1

Reescribir

1

como

1^{2}

= tan^{4} (x) - 1^{2}

Aplicar las leyes de los exponentes:

a^{b c} = (a^{b})^{c}

tan^{4} (x) = (tan^{2} (x))^{2}

= (tan^{2} (x))^{2} - 1^{2}

= (tan^{2} (x))^{2} - 1^{2}

Aplicar la siguiente regla para binomios al cuadrado:

x^{2} - y^{2} = (x + y) (x - y)

(tan^{2} (x))^{2} - 1^{2} = (tan^{2} (x) + 1) (tan^{2} (x) - 1)

= (tan^{2} (x) + 1) (tan^{2} (x) - 1)

Factorizar

tan^{2} (x) - 1 : (tan (x) + 1) (tan (x) - 1)

tan^{2} (x) - 1

Reescribir

1

como

1^{2}

= tan^{2} (x) - 1^{2}

Aplicar la siguiente regla para binomios al cuadrado:

x^{2} - y^{2} = (x + y) (x - y)

tan^{2} (x) - 1^{2} = (tan (x) + 1) (tan (x) - 1)

= (tan (x) + 1) (tan (x) - 1)

= (tan^{2} (x) + 1) (tan (x) + 1) (tan (x) - 1)

= - (tan^{2} (x) + 1) (tan (x) + 1) (tan (x) - 1)

= - \frac{( tan ^{2} ( x ) + 1 ) ( tan ( x ) + 1 ) ( tan ( x ) - 1 )}{1 - tan ^{2} ( x )}

Factorizar

1 - tan^{2} (x) : - (tan (x) + 1) (tan (x) - 1)

1 - tan^{2} (x)

Factorizar el termino común

- 1

= - (tan^{2} (x) - 1)

Factorizar

tan^{2} (x) - 1 : (tan (x) + 1) (tan (x) - 1)

tan^{2} (x) - 1

Reescribir

1

como

1^{2}

= tan^{2} (x) - 1^{2}

Aplicar la siguiente regla para binomios al cuadrado:

x^{2} - y^{2} = (x + y) (x - y)

tan^{2} (x) - 1^{2} = (tan (x) + 1) (tan (x) - 1)

= (tan (x) + 1) (tan (x) - 1)

= - (tan (x) + 1) (tan (x) - 1)

= \frac{( tan ^{2} ( x ) + 1 ) ( tan ( x ) + 1 ) ( tan ( x ) - 1 )}{( tan ( x ) + 1 ) ( tan ( x ) - 1 )}

Cancelar

\frac{( tan ^{2} ( x ) + 1 ) ( tan ( x ) + 1 ) ( tan ( x ) - 1 )}{( tan ( x ) + 1 ) ( tan ( x ) - 1 )} : tan^{2} (x) + 1

\frac{( tan ^{2} ( x ) + 1 ) ( tan ( x ) + 1 ) ( tan ( x ) - 1 )}{( tan ( x ) + 1 ) ( tan ( x ) - 1 )}

Eliminar los terminos comunes:

tan (x) + 1

= \frac{( tan ^{2} ( x ) + 1 ) ( tan ( x ) - 1 )}{tan ( x ) - 1}

Eliminar los terminos comunes:

tan (x) - 1

= tan^{2} (x) + 1

= tan^{2} (x) + 1

= 1 + tan^{2} (x)

Re-escribir usando identidades trigonométricas

1 + tan^{2} (x)

Utilizar la identidad pitagórica:

tan^{2} (x) + 1 = sec^{2} (x)

= sec^{2} (x)

= sec^{2} (x)

Se demostró que ambos lados pueden tomar la misma forma

\Rightarrow Verdadero

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