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

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x^{2} + y^{2} + z^{2} = 2 x y z

x^{2} + y^{2} + z^{2} + 2 x y z = 0

x y + y z + z x = x y z

x^{2} + y^{2} + z^{2} + 2 x y z = 1

answer is C.

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Detailed Solution

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

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

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

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

$∴ x^{2} + y^{2} + z^{2} + 2 x y z = 1$

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