Prove that 1 a a - 1 + b - 1 + 1 a a - 1 - b - 1 = 2 b 2 b + a b - a

Hi, LHS, 1aa-1+b-1+1aa-1-b-1=1a1a+1b+1a1a-1b=1aa+bab+1ab-aab=1a×aba+b+1a×abb-a= ba+b+bb-a= b1a+b+1b-a=bb-a+a+ba+bb-a= 2b2a+bb-a= RHS

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

Hi, LHS, \frac{\frac{1}{a}}{a^{- 1} + b^{- 1}} + \frac{\frac{1}{a}}{a^{- 1} - b^{- 1}} = \frac{\frac{1}{a}}{\frac{1}{a} + \frac{1}{b}} + \frac{\frac{1}{a}}{\frac{1}{a} - \frac{1}{b}} = \frac{\frac{1}{a}}{\frac{a + b}{ab}} + \frac{\frac{1}{a}}{\frac{b - a}{ab}} = \frac{1}{a} \times \frac{ab}{a + b} + \frac{1}{a} \times \frac{ab}{b - a} = \frac{b}{a + b} + \frac{b}{b - a} = b [\frac{1}{a + b} + \frac{1}{b - a}] = b [\frac{b - a + a + b}{(a + b) (b - a)}] = \frac{2 b^{2}}{(a + b) (b - a)} = RHS