Question

Comprobar las siguientes identidades trigonométricas:

tgα+cotα=secα+cosecα

(senα+cotα)/(tgα+cosc)=cosα

Answer 1

Respuesta:

a)

$tg(\alpha)+cot(\alpha)=sec(\alpha)+cosec(\alpha)\\\\\frac{sen(\alpha)}{cos(\alpha)} +\frac{cos(\alpha)}{sen(\alpha)} =sec(\alpha)+cosec(\alpha)\\\\\frac{sen^2(\alpha)+ cos^2(\alpha)}{sen(\alpha).cos(\alpha) } =sec(\alpha)+cosec(\alpha)$

Sabemos que una de las identidades trigonométricas indica que:

$sen^2(\alpha)+ cos^2(\alpha)=1$

Por lo tanto:

$\frac{1}{sen(\alpha).cos(\alpha) } =sec(\alpha)+cosec(\alpha) \\\\\frac{1}{sen(\alpha)}+\frac{1}{cos(\alpha) } =sec(\alpha)+cosec(\alpha) \\\\cosec(\alpha)+sec(\alpha) =sec(\alpha)+cosec(\alpha)$

b)

$\frac{sen(\alpha)+cot(\alpha) }{tga(\alpha)+cosc(\alpha)} =cos(\alpha)\\\\(sen(\alpha)+\frac{cos(\alpha)}{sen(\alpha)})/(\frac{sen(\alpha)}{cos(\alpha)}+\frac{1}{sen(\alpha)} )=cos(\alpha )\\\\(\frac{sen^2(\alpha)+cos(\alpha)}{sen(\alpha)})/(\frac{sen^2(\alpha)+cos(\alpha)}{sen(\alpha).cos(\alpha)} )=cos(\alpha)\\\\(\frac{sen^2(\alpha)+cos(\alpha)}{sen(\alpha)}).(\frac{sen(\alpha).cos(\alpha)}{sen^2(\alpha)+cos(\alpha)} )=cos(\alpha)$

$\frac{sen(\alpha).cos(\alpha)}{sen(\alpha)} =cos(\alpha)\\\\cos(\alpha)=cos(\alpha)$

Espero que te sirva, saludos!