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

Solución

verificar

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

Verdadero

Pasos de solución

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

Manipular el lado derecho

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

Simplificar

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

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

Factorizar

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

- 1 + csc^{4} (x)

Reescribir

csc^{4} (x) - 1

como

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

csc^{4} (x) - 1

Reescribir

1

como

1^{2}

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

Aplicar las leyes de los exponentes:

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

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

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

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

Aplicar la siguiente regla para binomios al cuadrado:

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

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

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

Factorizar

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

csc^{2} (x) - 1

Reescribir

1

como

1^{2}

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

Aplicar la siguiente regla para binomios al cuadrado:

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

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

= (csc (x) + 1) (csc (x) - 1)

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

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

Eliminar los terminos comunes:

csc^{2} (x) + 1

= (csc (x) + 1) (csc (x) - 1)

= (- 1 + csc (x)) (1 + csc (x))

Re-escribir usando identidades trigonométricas

(- 1 + csc (x)) (1 + csc (x))

Expandir

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

(csc (x) + 1) (csc (x) - 1)

Aplicar la siguiente regla para binomios al cuadrado:

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

a = csc (x), b = 1

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

Aplicar la regla

1^{a} = 1

1^{2} = 1

= csc^{2} (x) - 1

= csc^{2} (x) - 1

Utilizar la identidad pitagórica:

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

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

= cot^{2} (x)

= cot^{2} (x)

Se demostró que ambos lados pueden tomar la misma forma

\Rightarrow Verdadero

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

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

sin^{2} (θ) + sin^{2} (θ) tan^{2} (θ) = tan^{2} (θ)

1 + sin (2 θ) = (cos (θ) sin (θ))^{2}

sec (θ) csc (θ) = tan (θ) + cot (θ)