Question

If ${\cos }^{-1}x+{\cos }^{-1}y+{\cos }^{-1}z=\pi$, prove that $x^2+y^2+z^2+2xyz=1$

Answer 1

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