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Popular >

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

Solución

verificar

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

Solución

Verdadero

Pasos de solución

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

Manipular el lado derecho

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

Re-escribir usando identidades trigonométricas

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

Utilizar la identidad pitagórica:

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

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

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

Simplificar

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

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

Factorizar

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

sec^{4} (x) - 1

Reescribir

sec^{4} (x) - 1

como

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

sec^{4} (x) - 1

Reescribir

1

como

1^{2}

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

Aplicar las leyes de los exponentes:

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

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

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

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

Aplicar la siguiente regla para binomios al cuadrado:

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

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

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

Factorizar

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

sec^{2} (x) - 1

Reescribir

1

como

1^{2}

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

Aplicar la siguiente regla para binomios al cuadrado:

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

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

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

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

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

Factorizar

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

sec^{2} (x) - 1

Reescribir

1

como

1^{2}

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

Aplicar la siguiente regla para binomios al cuadrado:

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

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

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

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

Cancelar

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

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

Eliminar los terminos comunes:

sec (x) + 1

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

Eliminar los terminos comunes:

sec (x) - 1

= sec^{2} (x) + 1

= sec^{2} (x) + 1

= sec^{2} (x) + 1

Utilizar la identidad pitagórica:

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

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

Simplificar

= tan^{2} (x) + 2

= tan^{2} (x) + 2

Se demostró que ambos lados pueden tomar la misma forma

\Rightarrow Verdadero

Ejemplos populares

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

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tan (x) + cot (x) = \frac{1}{sin ( x ) cos ( x )}

verificar (cot^2(x))/(csc(x)-1)=csc(x)+1 verificar

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

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

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

verificar tan^2(θ)-sin^2(θ)=sin^4(θ)sec^2(θ)verificar

tan^{2} (θ) - sin^{2} (θ) = sin^{4} (θ) sec^{2} (θ)