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

Solución

verificar

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

Verdadero

Pasos de solución

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

Manipular el lado derecho

tan^{4} (x) - sec^{4} (x)

Factorizar

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

- sec^{4} (x) + tan^{4} (x)

Reescribir

tan^{4} (x) - sec^{4} (x)

como

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

tan^{4} (x) - sec^{4} (x)

Aplicar las leyes de los exponentes:

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

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

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

Aplicar las leyes de los exponentes:

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

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

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

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

Aplicar la siguiente regla para binomios al cuadrado:

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

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

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

Factorizar

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

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

Aplicar la siguiente regla para binomios al cuadrado:

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

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

= (tan (x) + sec (x)) (tan (x) - sec (x))

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

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

Re-escribir usando identidades trigonométricas

(- sec (x) + tan (x)) (sec (x) + tan (x)) (sec^{2} (x) + tan^{2} (x))

Utilizar la identidad pitagórica:

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

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

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

Sumar elementos similares:

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

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

Expandir

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

(tan (x) + sec (x)) (tan (x) - sec (x))

Aplicar la siguiente regla para binomios al cuadrado:

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

a = tan (x), b = sec (x)

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

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

Utilizar la identidad pitagórica:

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

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

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

Simplificar

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

(- 1 + 2 sec^{2} (x)) (- 1)

Quitar los parentesis:

(- a) = - a

= - (- 1 + 2 sec^{2} (x)) \cdot 1

Multiplicar:

(- 1 + 2 sec^{2} (x)) \cdot 1 = (- 1 + 2 sec^{2} (x))

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

Poner los parentesis

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

Aplicar las reglas de los signos

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

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

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

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

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

Se demostró que ambos lados pueden tomar la misma forma

\Rightarrow Verdadero

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

(cos (x) + sin (x))^{2} - 2 sin (x) cos (x) = 1

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

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

8 csc^{2} (x) - 3 cot^{2} (x) = 3 + 5 csc^{2} (x)