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Popular Trigonometría >

verificar-(-1+sin^2(θ))/(sin^2(θ))=cot^2(θ)

Solución

verificar

- \frac{- 1 + sin ^{2} ( θ )}{sin ^{2} ( θ )} = cot^{2} (θ)

Solución

Verdadero

Pasos de solución

- \frac{- 1 + sin ^{2} ( θ )}{sin ^{2} ( θ )} = cot^{2} (θ)

Manipular el lado izquierdo

cot^{2} (θ)

Expresar con seno, coseno

cot^{2} (θ)

Utilizar la identidad trigonométrica básica:

cot (x) = \frac{cos ( x )}{sin ( x )}

= (\frac{cos ( θ )}{sin ( θ )})^{2}

Simplificar

(\frac{cos ( θ )}{sin ( θ )})^{2} : \frac{cos ^{2} ( θ )}{sin ^{2} ( θ )}

(\frac{cos ( θ )}{sin ( θ )})^{2}

Aplicar las leyes de los exponentes:

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

= \frac{cos ^{2} ( θ )}{sin ^{2} ( θ )}

= \frac{cos ^{2} ( θ )}{sin ^{2} ( θ )}

= \frac{cos ^{2} ( θ )}{sin ^{2} ( θ )}

Re-escribir usando identidades trigonométricas

\frac{cos ^{2} ( θ )}{sin ^{2} ( θ )}

Utilizar la identidad trigonométrica del ángulo doble:

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

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

= \frac{cos ( 2 θ ) + sin ^{2} ( θ )}{sin ^{2} ( θ )}

Utilizar la identidad pitagórica:

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

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

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

Utilizar la identidad trigonométrica del ángulo doble:

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

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

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

cos^{2} (θ) - sin^{2} (θ) + 1 - cos^{2} (θ)

Agrupar términos semejantes

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

Sumar elementos similares:

cos^{2} (θ) - cos^{2} (θ) = 0

= - sin^{2} (θ) + 1

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

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

Factorizar

\frac{1 - sin ^{2} ( θ )}{1 - cos ^{2} ( θ )} : \frac{- ( sin ( θ ) + 1 ) ( sin ( θ ) - 1 )}{- ( cos ( θ ) + 1 ) ( cos ( θ ) - 1 )}

\frac{1 - sin ^{2} ( θ )}{1 - cos ^{2} ( θ )}

Factorizar

1 - cos^{2} (θ) : - (cos (θ) + 1) (cos (θ) - 1)

1 - cos^{2} (θ)

Factorizar el termino común

- 1

= - (cos^{2} (θ) - 1)

Factorizar

cos^{2} (θ) - 1 : (cos (θ) + 1) (cos (θ) - 1)

cos^{2} (θ) - 1

Reescribir

1

como

1^{2}

= cos^{2} (θ) - 1^{2}

Aplicar la siguiente regla para binomios al cuadrado:

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

cos^{2} (θ) - 1^{2} = (cos (θ) + 1) (cos (θ) - 1)

= (cos (θ) + 1) (cos (θ) - 1)

= - (cos (θ) + 1) (cos (θ) - 1)

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

Factorizar

1 - sin^{2} (θ) : - (sin (θ) + 1) (sin (θ) - 1)

1 - sin^{2} (θ)

Factorizar el termino común

- 1

= - (sin^{2} (θ) - 1)

Factorizar

sin^{2} (θ) - 1 : (sin (θ) + 1) (sin (θ) - 1)

sin^{2} (θ) - 1

Reescribir

1

como

1^{2}

= sin^{2} (θ) - 1^{2}

Aplicar la siguiente regla para binomios al cuadrado:

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

sin^{2} (θ) - 1^{2} = (sin (θ) + 1) (sin (θ) - 1)

= (sin (θ) + 1) (sin (θ) - 1)

= - (sin (θ) + 1) (sin (θ) - 1)

= \frac{- ( sin ( θ ) + 1 ) ( sin ( θ ) - 1 )}{- ( cos ( θ ) + 1 ) ( cos ( θ ) - 1 )}

= \frac{- ( - 1 + sin ( θ ) ) ( 1 + sin ( θ ) )}{- ( - 1 + cos ( θ ) ) ( 1 + cos ( θ ) )}

Expandir

\frac{- ( - 1 + sin ( θ ) ) ( 1 + sin ( θ ) )}{- ( - 1 + cos ( θ ) ) ( 1 + cos ( θ ) )} : \frac{sin ^{2} ( θ ) - 1}{cos ^{2} ( θ ) - 1}

\frac{- ( - 1 + sin ( θ ) ) ( 1 + sin ( θ ) )}{- ( - 1 + cos ( θ ) ) ( 1 + cos ( θ ) )}

Aplicar las propiedades de las fracciones:

\frac{- a}{- b} = \frac{a}{b}

= \frac{( - 1 + sin ( θ ) ) ( 1 + sin ( θ ) )}{( - 1 + cos ( θ ) ) ( 1 + cos ( θ ) )}

Expandir

(- 1 + cos (θ)) (1 + cos (θ)) : cos^{2} (θ) - 1

(- 1 + cos (θ)) (1 + cos (θ))

Aplicar la siguiente regla para binomios al cuadrado:

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

a = cos (θ), b = 1

= cos^{2} (θ) - 1^{2}

Aplicar la regla

1^{a} = 1

1^{2} = 1

= cos^{2} (θ) - 1

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

Expandir

(- 1 + sin (θ)) (1 + sin (θ)) : sin^{2} (θ) - 1

(- 1 + sin (θ)) (1 + sin (θ))

Aplicar la siguiente regla para binomios al cuadrado:

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

a = sin (θ), b = 1

= sin^{2} (θ) - 1^{2}

Aplicar la regla

1^{a} = 1

1^{2} = 1

= sin^{2} (θ) - 1

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

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

Re-escribir usando identidades trigonométricas

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

Utilizar la identidad pitagórica:

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

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

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

Simplificar

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

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

Se demostró que ambos lados pueden tomar la misma forma

\Rightarrow Verdadero

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