If $${\cos ^{ - 1}}x + {\cos ^{ - 1}}y + {\cos ^{ - 1}}z = \pi ,{\text{ then}}$$

Solution :
Given that,
$$\eqalign{ & {\cos ^{ - 1}}\left( x \right) + {\cos ^{ - 1}}\left( y \right) + {\cos ^{ - 1}}\left( z \right) = \pi \cr & \Rightarrow {\cos ^{ - 1}}\left( x \right) + {\cos ^{ - 1}}\left( y \right) + {\cos ^{ - 1}}\left( z \right) = {\cos ^{ - 1}}\left( { - 1} \right) \cr & \Rightarrow {\cos ^{ - 1}}\left( x \right) + {\cos ^{ - 1}}\left( y \right) = \pi - {\cos ^{ - 1}}\left( z \right) \cr & \Rightarrow {\cos ^{ - 1}}\left( {xy - \sqrt {1 - {x^2}} \sqrt {1 - {y^2}} } \right) = {\cos ^{ - 1}}\left( { - z} \right) \cr & \Rightarrow xy - \sqrt {\left( {1 - {x^2}} \right)\left( {1 - {y^2}} \right)} = - z \cr & \Rightarrow \left( {xy + z} \right) = \sqrt {\left( {1 - {x^2}} \right)\left( {1 - {y^2}} \right)} \cr} $$
Squaring both sides, we get
$${x^2} + {y^2} + {z^2} + 2xyz = 1.$$