When simplified ${(x^{- 1} + y^{- 1})}^{- 1}$ is equal to:

x + y

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\frac{x y}{x + y}

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x y

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\frac{1}{x y}

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Solution

The correct option is B $\frac{x y}{x + y}$
${(x^{- 1} + y^{- 1})}^{- 1} = {(\frac{1}{x} + \frac{1}{y})}^{- 1} = {(\frac{x + y}{x y})}^{- 1} = \frac{x y}{x + y}$

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Similar questions

When simplified $(x^{- 1} + y^{- 1})^{- 1}$ is equal to

[\frac{x + y}{x^{- 1} + y^{- 1}}] - x y

x y [(x + y) (\frac{1}{x} + \frac{1}{y}) - 2]

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

x = s e c ϕ - t a n ϕ

y = c o s e c ϕ + c o t ϕ

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