Question

Demostrar la siguiente identidad trigonométrica con proceso completo ​

Answer 1

Explicación paso a paso:

$\frac{1}{1-sen^{2}\alpha } +\frac{1}{1+csc^{2}\alpha } =1\\\\\frac{1+csc^{2}\alpha+1-sen^{2}\alpha}{(1-sen^{2}\alpha)(1+csc^{2}\alpha)} =1\\\\\frac{2+csc^{2}\alpha-sen^{2}\alpha}{1+csc^{2}\alpha-sen^{2}\alpha+csc^{2}\alpha*sen^{2}\alpha} =1$

Usamos la reciproca:

$csc\alpha=\frac{1}{sen\alpha }$

$csc\alpha*sen\alpha =1$

Que es igual a:

$csc^{2}\alpha*sen^{2}\alpha =1^{2} \\\\csc^{2}\alpha*sen^{2}\alpha =1$

Entonces:

$\frac{2+csc^{2}\alpha-sen^{2}\alpha}{1+csc^{2}\alpha-sen^{2}\alpha+1} =1\\\\\frac{2+csc^{2}\alpha-sen^{2}\alpha}{2+csc^{2}\alpha-sen^{2}\alpha} =1$

Simplificando tenemos:

$1=1$