If cot-1x-cot-1y-cot-1z-2-x-y-z- 0 and xy- 1- then x-y-z is also equal to

The correct option is B xyz tan^ 1(1x)+tan^ 1(1y)=π2 cot^ 1(z) ⇒ nbsp;tan^ 1(x+yxy 1)=tan^ 1(z) Taking tan on both the sides, we get ...

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Range of Trigonometric Expressions

If cot-1x+c...

Question

If $c o t^{- 1} x + c o t^{- 1} y + c o t^{- 1} z = \frac{π}{2}, x, y, z > 0 a n d x y < 1, t h e n x + y + z$ is also equal to

\frac{1}{x} + \frac{1}{y} + \frac{1}{z}

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x y z

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x y + y z + z x

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n o n e o f t h e s e

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Similar questions

c o t^{- 1} x + c o t^{- 1} y + c o t^{- 1} z = \frac{π}{4}

x y + y z + z x + x + y + z =

x > y > z > 0

{cot}^{- 1} \frac{x y + 1}{x - y} + {cot}^{- 1} \frac{y z + 1}{y - z} + {cot}^{- 1} \frac{z x + 1}{z - x}

{cot}^{- 1} x + {cot}^{- 1} y + {cot}^{- 1} z = \frac{π}{2}, x, y, z > 0

x y < 1

x + y + z

x > y > z > 0

{cot}^{- 1} (\frac{1 + x y}{x - y}) + {cot}^{- 1} (\frac{1 + x y}{y - z}) + {cot}^{- 1} (\frac{1 - z x}{z - x}) = π

x > 0, y > 0, z > 0, x y + y z + z x < 1

{tan}^{- 1} + {tan}^{- 1} y + {tan}^{- 1} z = π

x + y + z =

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