Question

a raise to the power minus 1 upon a raise to the power minus 1 + b raised to the power minus 1 + a raise to the power minus one upon a raise to the power minus 1 minus b raised to the power minus 1 is equals to 2 b x to the power 2 upon B square minus A square ​

Answer 1

$\textbf{To prove:}$

$\dfrac{a^{-1}}{a^{-1}+b^{-1}}+\dfrac{a^{-1}}{a^{-1}-b^{-1}}=\dfrac{2\,b^2}{b^2-a^2}$

$\textbf{Solution:}$

$\text{Consider,}$

$\dfrac{a^{-1}}{a^{-1}+b^{-1}}+\dfrac{a^{-1}}{a^{-1}-b^{-1}}$

$=a^{-1}[\dfrac{1}{a^{-1}+b^{-1}}+\dfrac{1}{a^{-1}-b^{-1}}]$

$=\frac{1}{a}[\dfrac{1}{\frac{1}{a}+\frac{1}{b}}+\dfrac{1}{\frac{1}{a}-\frac{1}{b}}]$

$=\frac{1}{a}[\dfrac{1}{\frac{b+a}{ab}}+\dfrac{1}{\frac{b-a}{ab}}]$

$=\frac{1}{a}[\dfrac{ab}{b+a}+\dfrac{ab}{b-a}]$

$=\frac{ab}{a}[\dfrac{1}{b+a}+\dfrac{1}{b-a}]$

$=b[\dfrac{b-a+b+a}{(b+a)(b-a)}]$

$=b[\dfrac{2b}{b^2-a^2}]$

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

$\implies\boxed{\bf\,\dfrac{a^{-1}}{a^{-1}+b^{-1}}+\dfrac{a^{-1}}{a^{-1}-b^{-1}}=\dfrac{2\,b^2}{b^2-a^2}}$

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Answer 2

Answer:

I don't know I don't know I don't know