Question

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

Answer 1

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

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

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

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

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

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

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

Squaring both sides

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

$x^2{y}^2+z^2+2xyz=1-x^2-y^2+x^2{y}^2$

$x^2+y^2+z^2+2xyz=1$

Hence proved.