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Mathematics

Prove that:
a−1(a−1+b−1)+a−1(a−1−b−1)=2b2(b2−a2)\dfrac{a^{-1}}{(a^{-1} + b^{-1})} + \dfrac{a^{-1}}{(a^{-1} - b^{-1})} = \dfrac{2b^2}{(b^2 - a^2)}

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Answer

Given,

a−1(a−1+b−1)+a−1(a−1−b−1)=2b2(b2−a2)\dfrac{a^{-1}}{(a^{-1} + b^{-1})} + \dfrac{a^{-1}}{(a^{-1} - b^{-1})} = \dfrac{2b^2}{(b^2 - a^2)}

Solving L.H.S :

⇒a−1(a−1+b−1)+a−1(a−1−b−1)⇒1a(1a+1b)+1a(1a−1b)⇒1a×1(b+aab)+1a×1(b−aab)⇒1a×abb+a+1a×abb−a⇒bb+a+bb−a⇒b(b−a)+b(b+a)(b)2−(a)2⇒b2−ba+b2+bab2−a2⇒2b2b2−a2.\Rightarrow \dfrac{a^{-1}}{(a^{-1} + b^{-1})} + \dfrac{a^{-1}}{(a^{-1} - b^{-1})} \\[1em] \Rightarrow \dfrac{\dfrac{1}{a}}{\Big(\dfrac{1}{a} + \dfrac{1}{b}\Big)} + \dfrac{\dfrac{1}{a}}{\Big(\dfrac{1}{a} - \dfrac{1}{b}\Big)} \\[1em] \Rightarrow \dfrac{1}{a} \times \dfrac{1}{\Big(\dfrac{b + a}{ab}\Big)} + \dfrac{1}{a} \times \dfrac{1}{\Big(\dfrac{b - a}{ab}\Big)} \\[1em] \Rightarrow \dfrac{1}{a} \times \dfrac{ab}{b + a} + \dfrac{1}{a} \times \dfrac{ab}{b - a} \\[1em] \Rightarrow \dfrac{b}{b + a} + \dfrac{b}{b - a} \\[1em] \Rightarrow \dfrac{b(b - a) + b(b + a)}{(b)^2 - (a)^2} \\[1em] \Rightarrow \dfrac{b^2 - ba + b^2 + ba}{b^2 - a^2} \\[1em] \Rightarrow \dfrac{2b^2}{b^2 - a^2}.

Hence proved, a−1(a−1+b−1)+a−1(a−1−b−1)=2b2(b2−a2)\dfrac{a^{-1}}{(a^{-1} + b^{-1})} + \dfrac{a^{-1}}{(a^{-1} - b^{-1})} = \dfrac{2b^2}{(b^2 - a^2)}.

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