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

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xy+yz+zx=1

x2+y2+z2+2xyz=1

x+y+z=xyz

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Correct option is A

Let tan-1x=A, tan-1y=B and tan-1z=C ⇒tanA=x, tanB=y and tanC=zNow 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=π2 then

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