Prove the following:

(a + b)^-1(a^-1 + b^-1) = $\dfrac{1}{ab}$

Given,

$\Rightarrow (a + b)^{-1}(a^{-1} + b^{-1}) = \dfrac{1}{ab}$

Solving L.H.S. of above equation,

$\Rightarrow \dfrac{1}{a + b} \times \Big(\dfrac{1}{a} + \dfrac{1}{b}\Big) \\[1em] \Rightarrow \dfrac{1}{a + b} \times \Big(\dfrac{b + a}{ab}\Big) \\[1em] \Rightarrow \dfrac{1}{ab}.$

Since, L.H.S. = R.H.S.

Hence, proved that (a + b)^-1(a^-1 + b^-1) = $\dfrac{1}{ab}$.