Question

así sii
$\frac{ \tan(x) - \sin(x) }{ \sin^{3} (x) } = \frac{ \sec(x) }{1 + \cos(x) }$
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Answer 1

Respuesta:

$\boxed{\frac{tan(x)-sin(x)}{sin^3(x)} }\\\boxed{=\frac{\frac{sen(x)}{cos(x)} -sen(x)}{sen(x)(1-cos^2(x)) }}}\\\boxed{=\frac{\frac{sen(x)}{cos(x)} -sen(x)}{sen(x)(1-cos(x))(1+cos(x))}}\\\boxed{=\frac{\frac{sen(x)-cos(x)sen(x)}{cos(x)}}{sen(x)(1-cos(x))(1+cos(x))}}\\\boxed{=\frac{\frac{sen(x)(1-cos(x))}{cos(x)}}{sen(x)(1-cos(x))(1+cos(x))}}\\\boxed{=\frac{\frac{sin(x)}{cos(x)} }{sen(x)(1+cos(x))} }\\\boxed{=\frac{\frac{1}{cos(x)} }{(1+cos(x))}}\\\boxed{=\frac{sec(x) }{1+cos(x)}}$

Listo!!!!!!!

Answer 2

Recuerda:

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

$\sin {}^{2} (x) = 1 - \cos {}^{2} (x)$

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

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

...

$\frac{ \frac{ \sin(x) }{ \cos(x) } - \sin(x) }{ \sin {}^{3} (x) } = \frac{ \frac{1}{ \cos(x) } }{1 + \cos(x) }$

$\frac{ \sin(x) - \sin(x) \cos(x) }{ \cos(x) \sin {}^{3} (x)} = \frac{1}{ \cos(x)(1 + \cos(x) ) }$

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

$(1 - \cos(x) )(1 + \cos(x) ) = \sin {}^{2} (x)$

$1 - \cos {}^{2} (x) = \sin {}^{2} (x)$

$\sin {}^{2} (x) = \sin {}^{2} (x)$

$1 = 1$

La igualdad es verdadera