Question

Prove that
$\frac{a {}^{ - 1} }{a {}^{ - 1} + b {}^{ - 1} } + \frac{a {}^{ - 1} }{a {}^{ - 1} - b {}^{ - 1} } = \frac{2b {}^{2} }{b {}^{2} - a {}^{2} }$
with steps

Answer 1

HEYA !

To prove :

$\frac{a {}^{ - 1} }{a {}^{ - 1} + b {}^{ - 1} } + \frac{a {}^{ - 1} }{a {}^{ - 1} - b {}^{ - 1} } = \frac{2b {}^{2} }{b {}^{2} - a {}^{2} }$
Let's Start with LHS , We have

LHS

$=\frac{a {}^{ - 1} }{a {}^{ - 1} + b {}^{ - 1} }+ \frac{a {}^{ - 1} }{a {}^{ - 1} - b {}^{ - 1} } \\ \\ = \frac{ \frac{1}{a} }{ \frac{1}{a} + \frac{1}{b} } + \frac{ \frac{1}{a} }{ \frac{1}{a} - \frac{1}{b} } \: \: \: \: \: \: \biggl \{ \: \: \because {p}^{ - 1} = \frac{1}{p} \: \: \biggl\} \\ \\ = \frac{ \frac{1}{a} }{ \frac{b + a}{ab} } + \frac{ \frac{1}{a} }{ \frac{b - a}{ab} } \\ \\ = \frac{ab}{a(b + a)} + \frac{ab}{a(b - a)} \\ \\ = \frac{b}{b + a} + \frac{b}{b - a} \\ \\ = \frac{b(b - a) + b(b + a)}{(b + a)(b - a)} \\ \\ = \frac{ {b}^{2} - ab + {b}^{2} + ab}{ {b}^{2} - {a}^{2} } \\ \\ = \frac{2 {b}^{2} }{ {b}^{2} - {a}^{2} }$
= RHS [ Proved ]