English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

1+cot^2(a)=tan^2(a)

Solución

1 + cot^{2} (a) = tan^{2} (a)

Solución

a = 0.90455 \dots + πn, a = 2.23703 \dots + πn

Grados

a = 51.82729 \dots^{\circ} + 18 0^{\circ} n, a = 128.17270 \dots^{\circ} + 18 0^{\circ} n

Pasos de solución

1 + cot^{2} (a) = tan^{2} (a)

Restar

tan^{2} (a)

de ambos lados

1 + cot^{2} (a) - tan^{2} (a) = 0

Re-escribir usando identidades trigonométricas

1 + cot^{2} (a) - tan^{2} (a)

Utilizar la identidad trigonométrica básica:

tan (x) = \frac{1}{cot ( x )}

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

(\frac{1}{cot ( a )})^{2} = \frac{1}{cot ^{2} ( a )}

(\frac{1}{cot ( a )})^{2}

Aplicar las leyes de los exponentes:

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

= \frac{1 ^{2}}{cot ^{2} ( a )}

Aplicar la regla

1^{a} = 1

1^{2} = 1

= \frac{1}{cot ^{2} ( a )}

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

1 + cot^{2} (a) - \frac{1}{cot ^{2} ( a )} = 0

Usando el método de sustitución

1 + cot^{2} (a) - \frac{1}{cot ^{2} ( a )} = 0

Sea:

cot (a) = u

1 + u^{2} - \frac{1}{u ^{2}} = 0

1 + u^{2} - \frac{1}{u ^{2}} = 0 : u = \frac{- 1 + 5}{2}, u = - \frac{- 1 + 5}{2}, u = \frac{- 1 - 5}{2}, u = - \frac{- 1 - 5}{2}

1 + u^{2} - \frac{1}{u ^{2}} = 0

Multiplicar ambos lados por

u^{2}

1 + u^{2} - \frac{1}{u ^{2}} = 0

Multiplicar ambos lados por

u^{2}

1 \cdot u^{2} + u^{2} u^{2} - \frac{1}{u ^{2}} u^{2} = 0 \cdot u^{2}

Simplificar

1 \cdot u^{2} + u^{2} u^{2} - \frac{1}{u ^{2}} u^{2} = 0 \cdot u^{2}

Simplificar

1 \cdot u^{2} : u^{2}

1 \cdot u^{2}

Multiplicar:

1 \cdot u^{2} = u^{2}

= u^{2}

Simplificar

u^{2} u^{2} : u^{4}

u^{2} u^{2}

Aplicar las leyes de los exponentes:

a^{b} \cdot a^{c} = a^{b + c}

u^{2} u^{2} = u^{2 + 2}

= u^{2 + 2}

Sumar:

2 + 2 = 4

= u^{4}

Simplificar

- \frac{1}{u ^{2}} u^{2} : - 1

- \frac{1}{u ^{2}} u^{2}

Multiplicar fracciones:

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

= - \frac{1 \cdot u ^{2}}{u ^{2}}

Eliminar los terminos comunes:

u^{2}

= - 1

Simplificar

0 \cdot u^{2} : 0

0 \cdot u^{2}

Aplicar la regla

0 \cdot a = 0

= 0

u^{2} + u^{4} - 1 = 0

u^{2} + u^{4} - 1 = 0

u^{2} + u^{4} - 1 = 0

Resolver

u^{2} + u^{4} - 1 = 0 : u = \frac{- 1 + 5}{2}, u = - \frac{- 1 + 5}{2}, u = \frac{- 1 - 5}{2}, u = - \frac{- 1 - 5}{2}

u^{2} + u^{4} - 1 = 0

Escribir en la forma binómica

a_{n} x^{n} + \dots + a_{1} x + a = 0

u^{4} + u^{2} - 1 = 0

Re-escribir la ecuación con

v = u^{2}

v^{2} = u^{4}

v^{2} + v - 1 = 0

Resolver

v^{2} + v - 1 = 0 : v = \frac{- 1 + 5}{2}, v = \frac{- 1 - 5}{2}

v^{2} + v - 1 = 0

Resolver con la fórmula general para ecuaciones de segundo grado:

v^{2} + v - 1 = 0

Formula general para ecuaciones de segundo grado:

Para

a = 1, b = 1, c = - 1

v_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 1 \cdot ( - 1 )}{2 \cdot 1}

v_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 1 \cdot ( - 1 )}{2 \cdot 1}

1^{2} - 4 \cdot 1 \cdot (- 1) = 5

1^{2} - 4 \cdot 1 \cdot (- 1)

Aplicar la regla

1^{a} = 1

1^{2} = 1

= 1 - 4 \cdot 1 \cdot (- 1)

Aplicar la regla

- (- a) = a

= 1 + 4 \cdot 1 \cdot 1

Multiplicar los numeros:

4 \cdot 1 \cdot 1 = 4

= 1 + 4

Sumar:

1 + 4 = 5

= 5

v_{1, 2} = \frac{- 1 \pm 5}{2 \cdot 1}

Separar las soluciones

v_{1} = \frac{- 1 + 5}{2 \cdot 1}, v_{2} = \frac{- 1 - 5}{2 \cdot 1}

v = \frac{- 1 + 5}{2 \cdot 1} : \frac{- 1 + 5}{2}

\frac{- 1 + 5}{2 \cdot 1}

Multiplicar los numeros:

2 \cdot 1 = 2

= \frac{- 1 + 5}{2}

v = \frac{- 1 - 5}{2 \cdot 1} : \frac{- 1 - 5}{2}

\frac{- 1 - 5}{2 \cdot 1}

Multiplicar los numeros:

2 \cdot 1 = 2

= \frac{- 1 - 5}{2}

Las soluciones a la ecuación de segundo grado son:

v = \frac{- 1 + 5}{2}, v = \frac{- 1 - 5}{2}

v = \frac{- 1 + 5}{2}, v = \frac{- 1 - 5}{2}

Sustituir hacia atrás la

v = u^{2},

resolver para

u

Resolver

u^{2} = \frac{- 1 + 5}{2} : u = \frac{- 1 + 5}{2}, u = - \frac{- 1 + 5}{2}

u^{2} = \frac{- 1 + 5}{2}

Para

x^{2} = f (a)

las soluciones son

x = f (a), - f (a)

u = \frac{- 1 + 5}{2}, u = - \frac{- 1 + 5}{2}

Resolver

u^{2} = \frac{- 1 - 5}{2} : u = \frac{- 1 - 5}{2}, u = - \frac{- 1 - 5}{2}

u^{2} = \frac{- 1 - 5}{2}

Para

x^{2} = f (a)

las soluciones son

x = f (a), - f (a)

u = \frac{- 1 - 5}{2}, u = - \frac{- 1 - 5}{2}

Las soluciones son

u = \frac{- 1 + 5}{2}, u = - \frac{- 1 + 5}{2}, u = \frac{- 1 - 5}{2}, u = - \frac{- 1 - 5}{2}

u = \frac{- 1 + 5}{2}, u = - \frac{- 1 + 5}{2}, u = \frac{- 1 - 5}{2}, u = - \frac{- 1 - 5}{2}

Verificar las soluciones

Encontrar los puntos no definidos (singularidades):

u = 0

Tomar el(los) denominador(es) de

1 + u^{2} - \frac{1}{u ^{2}}

y comparar con cero

Resolver

u^{2} = 0 : u = 0

u^{2} = 0

Aplicar la regla

x^{n} = 0 \Rightarrow x = 0

u = 0

Los siguientes puntos no están definidos

u = 0

Combinar los puntos no definidos con las soluciones:

u = \frac{- 1 + 5}{2}, u = - \frac{- 1 + 5}{2}, u = \frac{- 1 - 5}{2}, u = - \frac{- 1 - 5}{2}

Sustituir en la ecuación

u = cot (a)

cot (a) = \frac{- 1 + 5}{2}, cot (a) = - \frac{- 1 + 5}{2}, cot (a) = \frac{- 1 - 5}{2}, cot (a) = - \frac{- 1 - 5}{2}

cot (a) = \frac{- 1 + 5}{2}, cot (a) = - \frac{- 1 + 5}{2}, cot (a) = \frac{- 1 - 5}{2}, cot (a) = - \frac{- 1 - 5}{2}

cot (a) = \frac{- 1 + 5}{2} : a = arccot \frac{- 1 + 5}{2} + πn

cot (a) = \frac{- 1 + 5}{2}

Aplicar propiedades trigonométricas inversas

cot (a) = \frac{- 1 + 5}{2}

Soluciones generales para

cot (a) = \frac{- 1 + 5}{2}

cot (x) = a \Rightarrow x = arccot (a) + πn

a = arccot \frac{- 1 + 5}{2} + πn

a = arccot \frac{- 1 + 5}{2} + πn

cot (a) = - \frac{- 1 + 5}{2} : a = arccot - \frac{- 1 + 5}{2} + πn

cot (a) = - \frac{- 1 + 5}{2}

Aplicar propiedades trigonométricas inversas

cot (a) = - \frac{- 1 + 5}{2}

Soluciones generales para

cot (a) = - \frac{- 1 + 5}{2}

cot (x) = - a \Rightarrow x = arccot (- a) + πn

a = arccot - \frac{- 1 + 5}{2} + πn

a = arccot - \frac{- 1 + 5}{2} + πn

cot (a) = \frac{- 1 - 5}{2} : a = arccot \frac{- 1 - 5}{2} + πn

cot (a) = \frac{- 1 - 5}{2}

Aplicar propiedades trigonométricas inversas

cot (a) = \frac{- 1 - 5}{2}

Soluciones generales para

cot (a) = \frac{- 1 - 5}{2}

cot (x) = a \Rightarrow x = arccot (a) + πn

a = arccot \frac{- 1 - 5}{2} + πn

a = arccot \frac{- 1 - 5}{2} + πn

cot (a) = - \frac{- 1 - 5}{2} : a = arccot - \frac{- 1 - 5}{2} + πn

cot (a) = - \frac{- 1 - 5}{2}

Aplicar propiedades trigonométricas inversas

cot (a) = - \frac{- 1 - 5}{2}

Soluciones generales para

cot (a) = - \frac{- 1 - 5}{2}

cot (x) = a \Rightarrow x = arccot (a) + πn

a = arccot - \frac{- 1 - 5}{2} + πn

a = arccot - \frac{- 1 - 5}{2} + πn

Combinar toda las soluciones

a = arccot \frac{- 1 + 5}{2} + πn, a = arccot - \frac{- 1 + 5}{2} + πn, a = arccot \frac{- 1 - 5}{2} + πn, a = arccot - \frac{- 1 - 5}{2} + πn

Siendo que la ecuación esta indefinida para:

arccot \frac{- 1 - 5}{2} + πn, arccot - \frac{- 1 - 5}{2} + πn

a = arccot \frac{- 1 + 5}{2} + πn, a = arccot - \frac{- 1 + 5}{2} + πn

Mostrar soluciones en forma decimal

a = 0.90455 \dots + πn, a = 2.23703 \dots + πn

Ejemplos populares

tan(α)=(12)/(6sqrt(3))

tan (α) = \frac{12}{6 3}

1=cos^2(α)-2cos(α)sin(a)+sin^2(α)

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

4sin(θ)=1

4 sin (θ) = 1

sin(x+pi/2)=0.6

sin (x + \frac{π}{2}) = 0.6

4cot(x)+1=1+2cot(x)

4 cot (x) + 1 = 1 + 2 cot (x)