Let
[tex]\theta=\sin^{-1}\dfrac x2\implies\sin\theta=\dfrac x2[/tex]
Recall that
[tex]\cos\theta=\sqrt{1-\sin^2\theta}=\sqrt{1-\dfrac{x^2}4}[/tex]
Then
[tex]\tan\theta=\dfrac{\sin\theta}{\cos\theta}\implies\tan\left(\sin^{-1}\dfrac x2\right)=\dfrac{\frac x2}{\sqrt{1-\frac{x^2}4}}[/tex]
[tex]\implies\tan\left(\sin^{-1}\dfrac x2\right)=\dfrac x{\sqrt{4-x^2}}[/tex]
