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Parametric Differentiation

If 1 x - 1 ...

Question

If $\frac{1}{x} - \frac{1}{y} = \frac{1}{z}$ , then $z$ is equal to

A

y - x

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B

x - y

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C

\frac{y - x}{x y}

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D

\frac{x y}{y - x}

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E

\frac{x y}{x - y}

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Solution

The correct option is D $\frac{x y}{y - x}$
If $\frac{1}{x} - \frac{1}{y} = \frac{1}{z}$

$\Rightarrow \frac{1}{z} = \frac{y - x}{x y}$

$\Rightarrow z = \frac{x y}{y - x}$

option D is correct

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Similar questions

{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}}} =

x + y + z = x y z

\frac{x + y}{1 - x y} + \frac{y + z}{1 - y z} + \frac{z + x}{1 - z x} = \frac{x + y}{1 - x y} \cdot \frac{y + z}{1 - y z} \cdot \frac{z + x}{1 - z x}

\frac{x}{y + z - x} = \frac{y}{z + x - y} = \frac{z}{x + y - z}

\neq

0 then show that each ratio is equal to 1.

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

A . P .,

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

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