verificar csc(x)-cot(x)= 1/(csc(x)+cot(x))

Solución

verificar

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

Verdadero

Pasos de solución

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

Manipular el lado izquierdo

\frac{1}{csc ( x ) + cot ( x )}

Re-escribir usando identidades trigonométricas

\frac{1}{csc ( x ) + cot ( x )}

Utilizar la identidad pitagórica:

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

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

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

Simplificar

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

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

Aplicar la siguiente regla para binomios al cuadrado:

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

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

= \frac{( csc ( x ) + cot ( x ) ) ( csc ( x ) - cot ( x ) )}{csc ( x ) + cot ( x )}

Eliminar los terminos comunes:

csc (x) + cot (x)

= csc (x) - cot (x)

= csc (x) - cot (x)

= csc (x) - cot (x)

Se demostró que ambos lados pueden tomar la misma forma

\Rightarrow Verdadero

cos (π - θ) + sin (\frac{π}{2} + θ) = 0

sin (x) - csc (x) = - cos (x) cot (x)

tan (y) + cot (y) = sec (y) csc (y)

\frac{1}{sin ( x )} = csc (x)

\frac{csc ( θ )}{sec ( θ )} = cot (θ)