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(sec^2(A)+cot^2(A))/(csc^2(A))=sec^2(A)

Solución

\frac{sec ^{2} ( A ) + cot ^{2} ( A )}{csc ^{2} ( A )} = sec^{2} (A)

Solución

Sin soluci \overset{o}{ˊ} n para A \in R

Pasos de solución

\frac{sec ^{2} ( A ) + cot ^{2} ( A )}{csc ^{2} ( A )} = sec^{2} (A)

Restar

sec^{2} (A)

de ambos lados

\frac{sec ^{2} ( A ) + cot ^{2} ( A )}{csc ^{2} ( A )} - sec^{2} (A) = 0

Simplificar

\frac{sec ^{2} ( A ) + cot ^{2} ( A )}{csc ^{2} ( A )} - sec^{2} (A) : \frac{sec ^{2} ( A ) + cot ^{2} ( A ) - sec ^{2} ( A ) csc ^{2} ( A )}{csc ^{2} ( A )}

\frac{sec ^{2} ( A ) + cot ^{2} ( A )}{csc ^{2} ( A )} - sec^{2} (A)

Convertir a fracción:

sec^{2} (A) = \frac{sec ^{2} ( A ) csc ^{2} ( A )}{csc ^{2} ( A )}

= \frac{sec ^{2} ( A ) + cot ^{2} ( A )}{csc ^{2} ( A )} - \frac{sec ^{2} ( A ) csc ^{2} ( A )}{csc ^{2} ( A )}

Ya que los denominadores son iguales, combinar las fracciones:

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

= \frac{sec ^{2} ( A ) + cot ^{2} ( A ) - sec ^{2} ( A ) csc ^{2} ( A )}{csc ^{2} ( A )}

\frac{sec ^{2} ( A ) + cot ^{2} ( A ) - sec ^{2} ( A ) csc ^{2} ( A )}{csc ^{2} ( A )} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

sec^{2} (A) + cot^{2} (A) - sec^{2} (A) csc^{2} (A) = 0

Re-escribir usando identidades trigonométricas

cot^{2} (A) + sec^{2} (A) - csc^{2} (A) sec^{2} (A)

Utilizar la identidad pitagórica:

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

= cot^{2} (A) + sec^{2} (A) - (1 + cot^{2} (A)) sec^{2} (A)

Simplificar

cot^{2} (A) + sec^{2} (A) - (1 + cot^{2} (A)) sec^{2} (A) : cot^{2} (A) - sec^{2} (A) cot^{2} (A)

cot^{2} (A) + sec^{2} (A) - (1 + cot^{2} (A)) sec^{2} (A)

= cot^{2} (A) + sec^{2} (A) - sec^{2} (A) (1 + cot^{2} (A))

Expandir

- sec^{2} (A) (1 + cot^{2} (A)) : - sec^{2} (A) - sec^{2} (A) cot^{2} (A)

- sec^{2} (A) (1 + cot^{2} (A))

Poner los parentesis utilizando:

a (b + c) = ab + a c

a = - sec^{2} (A), b = 1, c = cot^{2} (A)

= - sec^{2} (A) \cdot 1 + (- sec^{2} (A)) cot^{2} (A)

Aplicar las reglas de los signos

+ (- a) = - a

= - 1 \cdot sec^{2} (A) - sec^{2} (A) cot^{2} (A)

Multiplicar:

1 \cdot sec^{2} (A) = sec^{2} (A)

= - sec^{2} (A) - sec^{2} (A) cot^{2} (A)

= cot^{2} (A) + sec^{2} (A) - sec^{2} (A) - sec^{2} (A) cot^{2} (A)

Sumar elementos similares:

sec^{2} (A) - sec^{2} (A) = 0

= cot^{2} (A) - sec^{2} (A) cot^{2} (A)

= cot^{2} (A) - sec^{2} (A) cot^{2} (A)

cot^{2} (A) - cot^{2} (A) sec^{2} (A) = 0

Factorizar

cot^{2} (A) - cot^{2} (A) sec^{2} (A) : - cot^{2} (A) (sec (A) + 1) (sec (A) - 1)

cot^{2} (A) - cot^{2} (A) sec^{2} (A)

Factorizar el termino común

- cot^{2} (A)

= - cot^{2} (A) (- 1 + sec^{2} (A))

Factorizar

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

sec^{2} (A) - 1

Reescribir

1

como

1^{2}

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

Aplicar la siguiente regla para binomios al cuadrado:

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

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

= (sec (A) + 1) (sec (A) - 1)

= - cot^{2} (A) (sec (A) + 1) (sec (A) - 1)

- cot^{2} (A) (sec (A) + 1) (sec (A) - 1) = 0

Resolver cada parte por separado

cot^{2} (A) = 0 or sec (A) + 1 = 0 or sec (A) - 1 = 0

cot^{2} (A) = 0 : A = \frac{π}{2} + πn

cot^{2} (A) = 0

Aplicar la regla

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

cot (A) = 0

Soluciones generales para

cot (A) = 0

cot (x)

tabla de valores periódicos con

πn

intervalos:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cot (x) \mp \infty 31 \frac{3}{3} 0 - \frac{3}{3} - 1 - 3

A = \frac{π}{2} + πn

A = \frac{π}{2} + πn

sec (A) + 1 = 0 : A = π + 2 πn

sec (A) + 1 = 0

Mover

1

al lado derecho

sec (A) + 1 = 0

Restar

1

de ambos lados

sec (A) + 1 - 1 = 0 - 1

Simplificar

sec (A) = - 1

sec (A) = - 1

Soluciones generales para

sec (A) = - 1

sec (x)

tabla de valores periódicos con

2 πn

intervalos:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sec (x) 1 \frac{2 3}{3} 22 Undefined - 2 - 2 - \frac{2 3}{3} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sec (x) - 1 - \frac{2 3}{3} - 2 - 2 Undefined 22 \frac{2 3}{3}

A = π + 2 πn

A = π + 2 πn

sec (A) - 1 = 0 : A = 2 πn

sec (A) - 1 = 0

Mover

1

al lado derecho

sec (A) - 1 = 0

Sumar

1

a ambos lados

sec (A) - 1 + 1 = 0 + 1

Simplificar

sec (A) = 1

sec (A) = 1

Soluciones generales para

sec (A) = 1

sec (x)

tabla de valores periódicos con

2 πn

intervalos:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sec (x) 1 \frac{2 3}{3} 22 Undefined - 2 - 2 - \frac{2 3}{3} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sec (x) - 1 - \frac{2 3}{3} - 2 - 2 Undefined 22 \frac{2 3}{3}

A = 0 + 2 πn

A = 0 + 2 πn

Resolver

A = 0 + 2 πn : A = 2 πn

A = 0 + 2 πn

0 + 2 πn = 2 πn

A = 2 πn

A = 2 πn

Combinar toda las soluciones

A = \frac{π}{2} + πn, A = π + 2 πn, A = 2 πn

Siendo que la ecuación esta indefinida para:

\frac{π}{2} + πn, π + 2 πn, 2 πn

Sin soluci \overset{o}{ˊ} n para A \in R

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