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

Question

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?

Answer 1

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|}}$ .