Show that $\sin^{-1}( x) =\tan^{-1}(x/\sqrt{1-x^2})$ for $|x| <1$ Show that $\sin^{-1}( x) = \tan^{-1}(x/\sqrt{1-x^2})$ for $|x| <1$
I have got as far as knowing that values of $\sin^{-1} x$ are only defined when x lies in the set [-1, 1].
and that for any value x, if $y = \sin^{-1}$ then $\sin y = x$.
But i am a little lost at how to use that to show this. Can anyone help me out here?
 A: Just for some variety, here's another method. If we construct a triangle from sections of your equation, a side derived from $1, x$ using pythagorus' theorem:

We can see from above that $A=\tan^{-1}{\frac{x}{\sqrt{1-x^2}}}$ and $A=\sin^{-1}{\frac{x}{1}}=\sin^{-1}{x}$.
Hence, as $(A=A)$ the statement is true.
A: So, $\displaystyle x=\sin y\implies -\frac\pi2\le y\le \frac\pi2$  (using the definition of principal value)
$\displaystyle\implies \cos y=+\sqrt{1-\sin^2y}=+\sqrt{1-x^2}, \tan y=\frac{\sin y}{\cos y}=\frac x{\sqrt{1-x^2}}$
$\displaystyle\implies \sin^{-1}x=\cos^{-1}\sqrt{1-x^2}=\tan^{-1}\frac x{\sqrt{1-x^2}} $
A: Hint: use two well-known identities:
\begin{align}
\tan \alpha &= \frac{\sin\alpha}{\cos\alpha} \\
\sin^2\alpha + \cos^2\alpha &= 1
\end{align}
and substitute $x=\sin\alpha$ to the original formula. You can do this since $\sin x$ is a bijection of $(-\frac{\pi}{2},+\frac{\pi}{2})$ onto $(-1,+1)$.
Spoiler:

 \begin{align} \sin^{−1} x &= \tan^{−1}\frac{x}{\sqrt{1−x^2}} \\ \sin^{−1}(\sin\alpha) &= \tan^{−1}\frac{\sin\alpha}{\sqrt{1−\sin^2\alpha}} \\ \alpha &= \tan^{−1}\frac{\sin\alpha}{\cos\alpha} \\ \alpha &= \tan^{−1}(\tan\alpha) \\ \alpha &= \alpha\end{align}

A: Let $\theta = \sin^{-1}(x)$, with $-\pi/2 < \theta < \pi/2$. Then $x = \sin \theta$, so $\sqrt{1-x^2} = \cos \theta$. Now it's easy!
A: Consider RHS, Put $x = \sin \alpha$, $\alpha =sin^{-1} x$.
By further solving, we get $\tan(\tan^{-1} \alpha)$ i.e $\alpha$, i.e $\sin^{-1} x$.
