Question

tgx/(1+sec⁡x )-tgx/(1-sec⁡x )=2/senx

Answer 1

Respuesta:ya va

Explicación paso a paso:

Answer 2

$\frac{ \tan(x) }{1 + \sec(x) } - \frac{ \tan(x) }{1 - \sec(x) } = \frac{2}{ \sin(x) } \\ \frac{ \ - tan(x) \sec(x) + \tan(x) - \tan(x) \tan(x) - \tan(x) }{1 - \sec(x) {}^{2} } = 2 \csc(x) \\ \frac{ - 2 \tan(x) \sec(x) }{ - \tan {}^{2} (x) } = \frac{2}{ \sin(x) } \\ \frac{ \frac{2}{ \cos(x) } }{ \frac{ \sin(x) }{ \cos(x) } } = \frac{2}{ \sin(x) } \\ \frac{2}{ \ \sin(x) } = \frac{2}{ \sin(x) }$

La expresion es igual.