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

We have,

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

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

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

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

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

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

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

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

On squaring both side and we get,

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

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

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

Hence proved.