If tan−1x+tan−1y+tan−1z=π2 and (x−y)2+(y−z)2+(z−x)2=0 then x2+y2+z2=

tan−1x+tan−1y+tan−1z=π2⇒tan−1[x+y+z−xyz1−xy−yz−zx]=π2 ⇒1−xy−yz−zx=0⇒xy+yz+zx=1 (x−y)2+(y−z)2+(z−x)2=x2+y2−2xy+y2+z2−2yz+z2+x2−2zx=0 ⇒2(x2+y2+z2)−2(xy+yz+zx)=0 ⇒x2+y2+z2=1

If tan−1x+tan−1y+tan−1z=π2 and (x−y)2+(y−z)2+(z−x)2=0 then x2+y2+z2=

tan−1x+tan−1y+tan−1z=π2⇒tan−1[x+y+z−xyz1−xy−yz−zx]=π2 ⇒1−xy−yz−zx=0⇒xy+yz+zx=1 (x−y)2+(y−z)2+(z−x)2=x2+y2−2xy+y2+z2−2yz+z2+x2−2zx=0 ⇒2(x2+y2+z2)−2(xy+yz+zx)=0 ⇒x2+y2+z2=1

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

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Detailed 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}

$\Rightarrow 1 - x y - y z - z x = 0 \Rightarrow x y + y z + z x = 1$

${(x - y)}^{2} + {(y - z)}^{2} + {(z - x)}^{2} = x^{2} + y^{2} - 2 x y + y^{2} + z^{2} - 2 y z + z^{2} + x^{2} - 2 z x = 0$ $\Rightarrow 2 (x^{2} + y^{2} + z^{2}) - 2 (x y + y z + z x) = 0$