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

${\cos }^{-1}x+{\cos }^{-1}y+{\cos }^{-1}z=\pi .......\left(1\right)$

$-1\le x,y,z\le 1$

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

Since

${\cos }^{-1}A+{\cos }^{-1}B={\cos }^{-1}[AB-\sqrt{1-A^2}.{\sqrt{1-B}}^2]$

where

$-1\le A,B\le 1$

therefore

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

$\cos \left\{{\cos }^{-1}[xy-\sqrt{1-x^2}.\sqrt{1-y^2}]\right\}=\cos \{\pi -{\cos }^{-1}z\}$

$xy-\sqrt{1-x^2}.\sqrt{1-y^2}=-\cos \left\{{\cos }^{-1}z\right\}[\because \cos \{\pi -\theta \}=-\cos \theta ]$

i.e

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

squaring on both sides :

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

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

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