The solution set of equation $\sin^{-1}{\sqrt{1-x^2}}\;+\cos^{-1}x=\cot^{-1}{\dfrac{\sqrt{1-x^2}}{x}}-\sin^{-1}x$ 
Find the solution set of equation $\sin^{-1}{\sqrt{1-x^2}}\;+\cos^{-1}x=\cot^{-1}{\dfrac{\sqrt{1-x^2}}{x}}-\sin^{-1}x$

My Approach:
$\sin^{-1}{\sqrt{1-x^2}}\;+\cos^{-1}x \; +\sin^{-1}x=\cot^{-1}{\dfrac{\sqrt{1-x^2}}{x}}$
$\implies$$\sin^{-1}{\sqrt{1-x^2}}\;+ \frac{\pi}{2}=\cot^{-1}{\dfrac{\sqrt{1-x^2}}{x}}$
$\implies$$\sin^{-1}{\sqrt{1-x^2}}\;+ \frac{\pi}{2}-\cot^{-1}{\dfrac{\sqrt{1-x^2}}{x}}=0$
$\implies$$\sin^{-1}{\sqrt{1-x^2}}\;+\tan^{-1}{\dfrac{\sqrt{1-x^2}}{x}}=0$
$\implies$$2\sin^{-1}{\sqrt{1-x^2}}=0$
I am obtaining two values of $x$ and those value are $x=1,{-1}.$
But given answer is $(-1,0)\cup \{1\}$.
What am I doing wrong?

The part where you went wrong is:

$\implies$$\sin^{-1}{\sqrt{1-x^2}}\;+\tan^{-1}{\dfrac{\sqrt{1-x^2}}{x}}=0$
$\implies$$2\sin^{-1}{\sqrt{1-x^2}}=0$

Because, $\sin^{-1}{\sqrt{1-x^2}}=$$\tan^{-1}{\dfrac{\sqrt{1-x^2}}{|x|}}$
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