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Popular Trigonometría >

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

Solución

verificar

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

Solución

Verdadero

Pasos de solución

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

Manipular el lado derecho

\frac{cos ( x )}{sec ( x ) - 1} - \frac{cos ( x )}{sec ( x ) + 1}

Expandir

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

\frac{cos ( x )}{- 1 + sec ( x )} - \frac{cos ( x )}{1 + sec ( x )}

Mínimo común múltiplo de

- 1 + sec (x), 1 + sec (x) : (sec (x) + 1) (sec (x) - 1)

- 1 + sec (x), 1 + sec (x)

Mínimo común múltiplo (MCM)

Calcular una expresión que este compuesta de factores que aparezcan tanto en

- 1 + sec (x)

1 + sec (x)

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

Reescribir las fracciones basandose en el mínimo común denominador

Multiplicar cada numerador por la misma cantidad necesaria para multiplicar el denominador correspondiente y convertirlo en el mínimo común denominador

Para

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

multiplicar el denominador y el numerador por

sec (x) + 1

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

Para

\frac{cos ( x )}{1 + sec ( x )} :

multiplicar el denominador y el numerador por

sec (x) - 1

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

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

Ya que los denominadores son iguales, combinar las fracciones:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

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

Expandir

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

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

Aplicar la siguiente regla para binomios al cuadrado:

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

a = sec (x), b = 1

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

Aplicar la regla

1^{a} = 1

1^{2} = 1

= sec^{2} (x) - 1

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

Expandir

cos (x) (sec (x) + 1) - cos (x) (sec (x) - 1) : 2 cos (x)

cos (x) (sec (x) + 1) - cos (x) (sec (x) - 1)

Expandir

cos (x) (sec (x) + 1) : cos (x) sec (x) + cos (x)

cos (x) (sec (x) + 1)

Poner los parentesis utilizando:

a (b + c) = ab + a c

a = cos (x), b = sec (x), c = 1

= cos (x) sec (x) + cos (x) \cdot 1

= cos (x) sec (x) + 1 \cdot cos (x)

Multiplicar:

1 \cdot cos (x) = cos (x)

= cos (x) sec (x) + cos (x)

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

Expandir

- cos (x) (sec (x) - 1) : - cos (x) sec (x) + cos (x)

- cos (x) (sec (x) - 1)

Poner los parentesis utilizando:

a (b - c) = ab - a c

a = - cos (x), b = sec (x), c = 1

= - cos (x) sec (x) - (- cos (x)) \cdot 1

Aplicar las reglas de los signos

- (- a) = a

= - cos (x) sec (x) + 1 \cdot cos (x)

Multiplicar:

1 \cdot cos (x) = cos (x)

= - cos (x) sec (x) + cos (x)

= cos (x) sec (x) + cos (x) - cos (x) sec (x) + cos (x)

Simplificar

cos (x) sec (x) + cos (x) - cos (x) sec (x) + cos (x) : 2 cos (x)

cos (x) sec (x) + cos (x) - cos (x) sec (x) + cos (x)

Sumar elementos similares:

cos (x) sec (x) - cos (x) sec (x) = 0

= cos (x) + cos (x)

Sumar elementos similares:

cos (x) + cos (x) = 2 cos (x)

= 2 cos (x)

= 2 cos (x)

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

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

Manipular el lado izquierdo

\frac{2 cos ( x )}{tan ^{2} ( x )}

Re-escribir usando identidades trigonométricas

\frac{2 cos ( x )}{tan ^{2} ( x )}

Utilizar la identidad pitagórica:

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

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

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

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

Se demostró que ambos lados pueden tomar la misma forma

\Rightarrow Verdadero

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