Question

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

Answer 1

Given,

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

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

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

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

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

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

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

Now squaring both sides we get,

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

or, $x^2+y^2+z^2+2xyz=1$.