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

Toolbox:
• $$cos^{-1}x+cos^{-1}y=cos^{-1} [ xy-\sqrt{1-x^2} \sqrt{1-y^2} ]$$
• $$cos^{-1}(-x)=\pi -cos^{-1}(x)$$
It is given that $$cos^{-1}x+cos^{-1}y+cos^{-1}z=\pi$$
$$\Rightarrow\: cos^{-1}x+cos^{-1}y=\pi-cos^{-1}z$$
Using the above formula of $$cos^{-1}x+cos^{-1}y$$ we get
$$\Rightarrow cos^{-1} [ xy-\sqrt{1-x^2} \sqrt{1-y^2} ] =cos^{-1}(-z)$$
Taking cos on both the sides we get
$$\Rightarrow xy-\sqrt{1-x^2} \sqrt{1-y^2}=cos(cos^{-1} (-z))=-z$$
$$\Rightarrow (xy+z)=\sqrt{1-x^2}.\sqrt{1-y^2}$$
Squaring on both the sides we get
$$x^2y^2+z^2+2xyz=(1-x^2)(1-y^2)=1+x^2y^2-x^2-y^2$$
Rearranging the terms we get
$$x^2+y^2+z^2+2xyz=1$$

Hence proved.

edited Mar 18, 2013