Question

What is tan(sin^-1(x/2))

Answer 1

Let

$\theta=\sin^{-1}\dfrac x2\implies\sin\theta=\dfrac x2$

Recall that

$\cos\theta=\sqrt{1-\sin^2\theta}=\sqrt{1-\dfrac{x^2}4}$

Then

$\tan\theta=\dfrac{\sin\theta}{\cos\theta}\implies\tan\left(\sin^{-1}\dfrac x2\right)=\dfrac{\frac x2}{\sqrt{1-\frac{x^2}4}}$

$\implies\tan\left(\sin^{-1}\dfrac x2\right)=\dfrac x{\sqrt{4-x^2}}$