Question

Prove that :

$\frac{a^{-1}}{(a{)}^{-1}+(b{)}^{-1}}+\frac{a^{-1}}{(a{)}^{-1}-(b{)}^{-1}}=\frac{2b^2}{b^2-a^2}$

Answer 1

$\frac{a^{-1}}{(a{)}^{-1}+(b{)}^{-1}}+\frac{a^{-1}}{(a{)}^{-1}-(b{)}^{-1}}=\frac{\frac{1}{a}}{\frac{1}{a}+\frac{1}{b}}+\frac{\frac{1}{a}}{\frac{1}{a}-\frac{1}{b}}[a^{-m}=(\frac{1}{a}{)}^m]$
$=\frac{\frac{1}{a}}{\frac{b+a}{ab}}+\frac{\frac{1}{a}}{\frac{b-a}{ab}}$

$=\frac{1}{a}.\frac{ab}{b+a}+\frac{1}{a}.\frac{ab}{b-a}$

$=\frac{b}{b+a}+\frac{b}{b-a}$

$=\frac{b(b-a)+b(b+a)}{(b+a)(b-a)}$

$=\frac{b^2-ba+b^2+ab}{b^2-a^2}$

$=\frac{2b^2}{b^2-a^2}$

Hence proved