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If $T a n^{- 1} x + T a n^{- 1} y + \tan^{- 1} z = \frac{π}{2}$ then

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xy+yz+zx=1

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Correct option is A

Let tan-1x=A, tan-1y=B, tan-1z=C. ⇒tanA=x, tanB=y, tanC=z. Given tan-1x+tan-1y+tan-1z=π2. ⇒A+B+C=π2 ⇒tanAtanB+tanBtanC+tanCtanA=1 ⇒xy+yz+zx=1

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If Tan−1x+Tan−1y+tan−1z=π2then

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If $T a n^{- 1} x + T a n^{- 1} y + \tan^{- 1} z = \frac{π}{2}$ then