Tan -1 x-tan -1 y-tan -1 z-2- 1-x y-y z-z x-

The correct option is B 0 tan^ 1 (x+y+z xyz/1 ( xy+yz+zx ) nbsp; )=π/2 ⇒1 ( xy+yz+zx )=0 ...

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First Fundamental Theorem of Calculus

tan -1 x+tan ...

Question

$t a n^{- 1} x + t a n^{- 1} y + t a n^{- 1} z = \frac{π}{2} \Rightarrow 1 - x y - y z - z x =$

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$I f t a n^{- 1} x + t a n^{- 1} y + t a n^{- 1} z = \frac{π}{2}, t h e n x y + y z + z x =$

If $t a n^{- 1} x + t a n^{- 1} y + t a n^{- 1} z = \frac{π}{2}$ then prove that $x y + y z + z x = 1$ .

x y + y z + z x = 1

{tan}^{- 1} x + {tan}^{- 1} y + {tan}^{- 1} z

{s i n}^{- 1} x, {c o s}^{- 1} x, {t a n}^{- 1} x

{t a n}^{- 1} x + {t a n}^{- 1} y = {t a n}^{- 1} \frac{x + y}{1 - x y}

x y < 1

π + {t a n}^{- 1} \frac{x + y}{1 - x y}

x y > 1

{c o s}^{- 1} p + {c o s}^{- 1} q + {c o s}^{- 1} r = π

p^{2} + q^{2} + r^{2} + 2 p q r = 1

{s i n}^{- 1} x + {s i n}^{- 1} y + {s i n}^{- 1} z = π

x^{4} + y^{4} + z^{4} + 4 x^{2} y^{2} z^{2} = 2 (x^{2} y^{2} + y^{2} z^{2} + z^{2} x^{2})

{t a n}^{- 1} x + {t a n}^{- 1} y + {t a n}^{- 1} z = π

π / 2

x + y + z = x y z

x y + y z + z x = 1

{t a n}^{- 1} x + {t a n}^{- 1} y = {t a n}^{- 1} \frac{x + y}{1 - x y}

x y < 1

= π + {t a n}^{- 1} \frac{x + y}{1 - x y}

x y > 1

{t a n}^{- 1} \sqrt{{\frac{x (x + y + z)}{y z}}} + {t a n}^{- 1} \sqrt{{\frac{y (x + y + z)}{z x}}} + {t a n}^{- 1} \sqrt{{\frac{z (x + y + z)}{x y}}} =

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