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Question

If $\tan^{- 1} x + \tan^{- 1} y + \tan^{- 1} z = \frac{π}{2}$ and ${(x - y)}^{2} + {(y - z)}^{2} + {(z - x)}^{2} = 0$ then $x^{2} + y^{2} + z^{2}$ is

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Solution

$\tan^{- 1} x + \tan^{- 1} y + \tan^{- 1} z = \frac{π}{2} \Rightarrow \tan^{- 1} (\frac{x + y + z - x y z}{1 - (x y + y z + z x)}) = \frac{π}{2} \Rightarrow \frac{x + y + z - x y z}{1 - (x y + y z + z x)} = \tan \frac{π}{2} = n o t d e f i n e d \Rightarrow 1 - (x y + y z + z x) = 0 \Rightarrow x y + y z + z x = 1 N o w, {(x - y)}^{2} + {(y - z)}^{2} + {(z - x)}^{2} = 0 \Rightarrow 2 (x^{2} + y^{2} + z^{2} - (x y + y z + z x)) = 0 \Rightarrow x^{2} + y^{2} + z^{2} - (x y + y z + z x) = 0 \Rightarrow x^{2} + y^{2} + z^{2} = 1$

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