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]