Respuesta :

Recall that

[tex]\cos^2\theta+\sin^2\theta=1[/tex]

which means

[tex]\dfrac{\sin\theta}{\sqrt{1-\sin^2\theta}}=\dfrac{\sin\theta}{\sqrt{\cos^2\theta}}[/tex]

Now, [tex]\cos\theta[/tex] could be positive or negative, which means [tex]\sqrt{\cos^2\theta}=|\cos\theta|[/tex]. If we specifically knew the sign of [tex]\cos\theta[/tex] was positive, then we can simplify and write

[tex]\dfrac{\sin\theta}{\sqrt{1-\sin^2\theta}}=\dfrac{\sin\theta}{\cos\theta}=\tan\theta[/tex]

or if it's negative,

[tex]\dfrac{\sin\theta}{\sqrt{1-\sin^2\theta}}=\dfrac{\sin\theta}{-\cos\theta}=-\tan\theta[/tex]

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