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Popular Trigonometría >

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

Solución

verificar

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

Solución

Verdadero

Pasos de solución

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

Manipular el lado derecho

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

Re-escribir usando identidades trigonométricas

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

Utilizar la identidad pitagórica:

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

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

Simplificar

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

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

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

Eliminar los terminos comunes:

tan^{2} (x) + 1

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

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

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

Expandir

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

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

Expandir

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

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

Aplicar la siguiente regla para binomios al cuadrado:

(a - b) (a + b) = a^{2} - b^{2}

a = tan (x), b = 1

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

Aplicar la regla

1^{a} = 1

1^{2} = 1

= tan^{2} (x) - 1

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

Poner los parentesis

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

Aplicar las reglas de los signos

- (- a) = a, - (a) = - a

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

= 1 - tan^{2} (x)

Se demostró que ambos lados pueden tomar la misma forma

\Rightarrow Verdadero

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