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العربية

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

Solución

verificar

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

Verdadero

Pasos de solución

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

Manipular el lado derecho

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

Simplificar

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

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

Factorizar

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

1 - 4 sec^{2} (x)

Factorizar el termino común

- 1

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

Factorizar

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

4 sec^{2} (x) - 1

Reescribir

4 sec^{2} (x) - 1

como

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

4 sec^{2} (x) - 1

Reescribir

4

como

2^{2}

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

Reescribir

1

como

1^{2}

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

Aplicar las leyes de los exponentes:

a^{m} b^{m} = (ab)^{m}

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

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

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

Aplicar la siguiente regla para binomios al cuadrado:

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

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

= (2 sec (x) + 1) (2 sec (x) - 1)

= - (2 sec (x) + 1) (2 sec (x) - 1)

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

Eliminar los terminos comunes:

2 sec (x) + 1

= - (2 sec (x) - 1)

Negar

- (2 sec (x) - 1) = - 2 sec (x) + 1

= - 2 sec (x) + 1

= 1 - 2 sec (x)

Se demostró que ambos lados pueden tomar la misma forma

\Rightarrow Verdadero