Well,
Given that [tex]x=\sin(\theta)[/tex],
We can rewrite the equation like,
[tex]\dfrac{\sin(\theta)}{\sqrt{1-\sin(\theta)^2}}[/tex]
Now use, [tex]\cos(\theta)^2+\sin(\theta)^2=1[/tex] which implies that [tex]1-\sin(\theta)^2=\cos(\theta)^2[/tex]
That means that,
[tex]\dfrac{\sin(\theta)}{\sqrt{1-\sin(\theta)^2}}\Longleftrightarrow\dfrac{\sin(\theta)}{\sqrt{\cos(\theta)^2}}[/tex]
By def [tex]\sqrt{x^2}=x[/tex] therefore [tex]\sqrt{\cos(\theta)^2}=\cos(\theta)[/tex]
So the fraction now looks like,
[tex]\dfrac{\sin(\theta)}{\cos(\theta)}[/tex]
Which is equal to the identity,
[tex]\boxed{\tan(\theta)}=\dfrac{\sin(\theta)}{\cos(\theta)}[/tex]
Hope this helps.
r3t40
