$\cos^{-1}x+\cos^{-1}y=\cos^{-1}\Big(xy-\sqrt{1-x^2}\sqrt{1-y^2}\Big)$ true for all $x$?

Question

$\cos^{-1}x+\cos^{-1}y=\cos^{-1}\Big(xy-\sqrt{1-x^2}\sqrt{1-y^2}\Big)$ true for all $x$ ?

My Attempt

Let $a=\cos^{-1}x$, $b=\cos^{-1}y\implies$$\cos a=x$, $\cos b=y$ and $a,b\in[0,\pi]\implies a+b\in[0,2\pi]$ $$ \cos(a+b)=\cos a\cos b-\sin a\sin b=xy-\sqrt{1-x^2}\sqrt{1-y^2}\\=\cos\bigg[\cos^{-1}\Big(xy-\sqrt{1-x^2}\sqrt{1-y^2}\Big)\bigg]\\ a+b=\color{red}{\cos^{-1}x+\cos^{-1}y=2n\pi\pm\cos^{-1}\Big(xy-\sqrt{1-x^2}\sqrt{1-y^2}\Big)} $$ Case 1: $a+b\in[0,\pi)$ $$ \cos^{-1}x+\cos^{-1}y=\cos^{-1}\Big(xy-\sqrt{1-x^2}\sqrt{1-y^2}\Big) $$

Case 2: $a+b\in[\pi,2\pi]$ $$ \cos^{-1}x+\cos^{-1}y=2\pi-\cos^{-1}\Big(xy-\sqrt{1-x^2}\sqrt{1-y^2}\Big) $$ Case 1-: $a+b\in[0,\pi)$

Note I have checked a possible solution to this in link, but it gives a different expression differ from my attempt.

Answer 1

Using Principal values and Why it's true? $\arcsin(x) +\arccos(x) = \frac{\pi}{2}$,

$\cos^{-1}x+\cos^{-1}y\le2\pi$

$$\cos^{-1}x+\cos^{-1}y=\cos^{-1}(xy-\sqrt{(1-x^2)(1-y^2)})$$

will hold true if $\cos^{-1}x+\cos^{-1}y\le\pi$

$\iff\sin^{-1}x+\sin^{-1}y\ge0\iff\sin^{-1}x\ge-\sin^{-1}y=\sin^{-1}(-y)$

$\iff x\ge -y\iff x+y\ge0$

$$\cos^{-1}x+\cos^{-1}y=2\pi-\cos^{-1}(xy-\sqrt{(1-x^2)(1-y^2)})$$

will hold true if $\cos^{-1}x+\cos^{-1}y>\pi$

$\iff\sin^{-1}x+\sin^{-1}y<0\iff x+y<0$