<s>[INST] Show that $\sin^{-1}x=\tan^{-1}\left(\frac{x}{\sqrt{1-x^2}}\right)$ Show that $\sin^{-1}x=\tan^{-1}\left(\frac{x}{\sqrt{1-x^2}}\right)$ for $|x|<1$
Question: what is wrong with the following reasoning?
Define: $x=\sin(y)$
$x^2=\sin^2(y)$
$1-x^2=\cos^2(y)$
$\frac{1}{1-x^2}=\sec^2(x)=1+\tan^2(y)$
$\tan^2(y)=\frac{1}{1-x^2}-1=\frac{x^2}{1-x^2}$
$\tan(y)=\pm\frac{|x|}{\sqrt{1-x^2}}$
$\sin^{-1}x=\tan^{-1}\left(\pm\frac{|x|}{\sqrt{1-x^2}}\right)$
So I end up with $\pm$ and the absolute-value-sign, which are not part of the original equation. How to get rid of them?
By the way, I know that there are other ways to show this statement is true. But the question is what is wrong with this method.
 [/INST] For example:
$$\begin{align*}&\left(\arcsin x\right)'=\frac1{\sqrt{1-x^2}}\\{}\\&\left(\arctan\frac x{\sqrt{1-x^2}}\right)'=\frac{\sqrt{1-x^2}+\frac{x^2}{\sqrt{1-x^2}}}{1-x^2}\cdot\frac1{1-\frac{x^2}{1-x^2}}=\frac1{\sqrt{1-x^2}}\end{align*}$$
and we know that if on some open interval $\;I\;$ we have $\;f'=g'\;$ then $\;f(x)=g(x)+K\;,\;\;K\;$ a constant, for all $\;x\in I\;$ , for for $\;x\in (-1,1)\;$
$$\arcsin x=\arctan\frac x{\sqrt{1-x^2}}+K$$
Now just choose $\;x=0\;$ to check $\;K=0\;$ and thus the equality
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