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verificar (sin(θ)+tan(θ))/(1+sec(θ))=sin(θ)

Solución

verificar

\frac{sin ( θ ) + tan ( θ )}{1 + sec ( θ )} = sin (θ)

Verdadero

Pasos de solución

\frac{sin ( θ ) + tan ( θ )}{1 + sec ( θ )} = sin (θ)

Manipular el lado derecho

\frac{sin ( θ ) + tan ( θ )}{1 + sec ( θ )}

Expresar con seno, coseno

\frac{sin ( θ ) + tan ( θ )}{1 + sec ( θ )}

Utilizar la identidad trigonométrica básica:

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

= \frac{sin ( θ ) + \frac{s i n ( θ )}{c o s ( θ )}}{1 + sec ( θ )}

Utilizar la identidad trigonométrica básica:

sec (x) = \frac{1}{cos ( x )}

= \frac{sin ( θ ) + \frac{s i n ( θ )}{c o s ( θ )}}{1 + \frac{1}{c o s ( θ )}}

Simplificar

\frac{sin ( θ ) + \frac{s i n ( θ )}{c o s ( θ )}}{1 + \frac{1}{c o s ( θ )}} : sin (θ)

\frac{sin ( θ ) + \frac{s i n ( θ )}{c o s ( θ )}}{1 + \frac{1}{c o s ( θ )}}

Simplificar

1 + \frac{1}{cos ( θ )}

en una fracción:

\frac{cos ( θ ) + 1}{cos ( θ )}

1 + \frac{1}{cos ( θ )}

Convertir a fracción:

1 = \frac{1 cos ( θ )}{cos ( θ )}

= \frac{1 \cdot cos ( θ )}{cos ( θ )} + \frac{1}{cos ( θ )}

Ya que los denominadores son iguales, combinar las fracciones:

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

= \frac{1 \cdot cos ( θ ) + 1}{cos ( θ )}

Multiplicar:

1 \cdot cos (θ) = cos (θ)

= \frac{cos ( θ ) + 1}{cos ( θ )}

= \frac{sin ( θ ) + \frac{s i n ( θ )}{c o s ( θ )}}{\frac{c o s ( θ ) + 1}{c o s ( θ )}}

Simplificar

sin (θ) + \frac{sin ( θ )}{cos ( θ )}

en una fracción:

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

sin (θ) + \frac{sin ( θ )}{cos ( θ )}

Convertir a fracción:

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

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

Ya que los denominadores son iguales, combinar las fracciones:

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

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

= \frac{\frac{s i n ( θ ) c o s ( θ ) + s i n ( θ )}{c o s ( θ )}}{\frac{c o s ( θ ) + 1}{c o s ( θ )}}

Dividir fracciones:

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

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

Eliminar los terminos comunes:

cos (θ)

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

Factorizar el termino común

sin (θ)

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

Eliminar los terminos comunes:

cos (θ) + 1

= sin (θ)

= sin (θ)

= sin (θ)

Se demostró que ambos lados pueden tomar la misma forma

\Rightarrow Verdadero

sec (2 θ) = \frac{sec ^{2} ( θ )}{1 - tan ^{2} ( θ )}

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

tan (θ) + cot (θ) = sec (θ) \cdot csc (θ)

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

\frac{cos ( x ) csc ( x )}{cot ^{2} ( x )} = tan (x)