Question

Prove that :

$\sf (i) \; \dfrac{ {a}^{-1}}{{a}^{-1} + {b}^{-1}} + \dfrac{{a}^{-1}}{{a}^{-1} - {b}^{-1}} = \dfrac{2b^2}{b^2 - a^2}$

$\sf (ii) \; \dfrac{a+b+c}{{a}^{-1}{b}^{-1}+{b}^{-1}{c}^{-1} + {c}^{-1}{a}^{-1}}=abc$​

Answer 1

Solution 1:

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

| We know that:

$\sf \Longrightarrow n^{-1} = \dfrac{1}{n}$

| Therefore:

$\sf \Longrightarrow \dfrac{\Bigg\{\dfrac{1}{a}\Bigg\}}{\Bigg\{\dfrac{1}{a} + \dfrac{1}{b} \Bigg\}} \ + \dfrac{\Bigg\{\dfrac{1}{a}\Bigg\}}{\Bigg\{\dfrac{1}{a} - \dfrac{1}{b} \Bigg\}}$

Taking LCM we get:

$\sf \Longrightarrow \dfrac{\Bigg\{\dfrac{1}{a}\Bigg\}}{\Bigg\{\dfrac{b + a}{ab}\Bigg\}} \ + \dfrac{\Bigg\{\dfrac{1}{a}\Bigg\}}{\Bigg\{\dfrac{b - a}{ab}\Bigg\}}$

$\sf \Longrightarrow \Bigg\{ \dfrac{1}{a} \times \dfrac{ab}{b + a} \Bigg\} + \Bigg\{ \dfrac{1}{a} \times \dfrac{ab}{b - a}\Bigg\}$

$\sf \Longrightarrow \Bigg\{\dfrac{b}{b + a} \Bigg\} + \Bigg\{ \dfrac{b}{b - a}\Bigg\}$

$\sf \Longrightarrow \dfrac{b\{b - a\} + b\{b + a\}}{\{b + a\}\{b - a\}}$

| Identities used:

(x + y)(x - y) = x² - y²

$\sf \Longrightarrow \dfrac{b^2 - ba + b^2 + ba}{b^2 - a^2}$

| ba and -ba get cancelled.

$\sf \Longrightarrow \dfrac{b^2 + b^2}{b^2 - a^2}$

$\sf \Longrightarrow \dfrac{2b^2}{b^2 - a^2}$

LHS = RHS

Hence proved.

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Solution 2:

$\sf \Longrightarrow \dfrac{a + b + c}{a^{-1}b^{-1} + b^{-1}c^{-1} + c^{-1}a^{-1}}$

| We know that:

$\sf \Longrightarrow n^{-1} = \dfrac{1}{n}$

| Therefore:

$\sf \Longrightarrow \dfrac{a + b + c}{\Bigg\{\dfrac{1}{a}\Bigg\}\Bigg\{\dfrac{1}{b}\Bigg\} + \Bigg\{\dfrac{1}{b}\Bigg\} \Bigg\{\dfrac{1}{c}\Bigg\} + \Bigg\{\dfrac{1}{c}\Bigg\}\Bigg\{\dfrac{1}{a}\Bigg\}}$

$\sf \Longrightarrow \dfrac{a + b + c}{\Bigg\{\dfrac{1}{ab}\Bigg\} + \Bigg\{\dfrac{1}{bc}\Bigg\} + \Bigg\{\dfrac{1}{ca}\Bigg\}}$

Taking LCM we get:

$\sf \Longrightarrow \dfrac{a + b + c}{\Bigg\{\dfrac{c + a + b}{abc}\Bigg\}}$

$\sf \Longrightarrow a + b + c \times \dfrac{abc}{c + a + b}$

(a + b + c) and (c + a + b) get cancelled.

$\sf \Longrightarrow abc$

LHS = RHS

Hence proved.

Answer 2

$\rule{170}2$

★ Solution 1:

$:\implies\sf \Bigg\lgroup\dfrac{a^{-1}}{a^{-1} + b^{-1}} + \dfrac{a^{-1}}{a^{-1} + b^{-1}}\Bigg\rgroup\\\\\\:\implies\sf \dfrac{\Bigg\lgroup\dfrac{1}{a}\Bigg\rgroup}{\Bigg\lgroup\dfrac{1}{a} + \dfrac{1}{b} \Bigg\rgroup} \ + \dfrac{\Bigg\lgroup\dfrac{1}{a}\Bigg\rgroup}{\Bigg\lgroup\dfrac{1}{a} - \dfrac{1}{b} \Bigg\rgroup}\\\\\\:\implies\sf \Bigg\lgroup\dfrac{1}{a}\times \dfrac{ab}{b + a} + \dfrac{1}{a}\times \dfrac{ab}{b - a}\Bigg\rgroup\\\\\\:\implies\sf \Bigg\lgroup\dfrac{b}{b + a} + \dfrac{b}{b - a}\Bigg\rgroup\\\\\\:\implies\sf \dfrac{b \Bigg\lgroup b - a\Bigg\rgroup + b \Bigg\lgroup b + a\Bigg\rgroup}{\Bigg\lgroup b + a\Bigg\rgroup\Bigg\lgroup b - a\Bigg\rgroup}\\\\\\:\implies\sf \dfrac{b^2 - ab + b^2 + ab}{b^2 - a^2}\\\\\\:\implies\boxed{\boxed{\bf \dfrac{2b^2}{b^2 - a^2}}}\:\bigstar\\\\\\ \underbrace{\red{\bf{Hence Proved}} }$

$\rule{150}1$

★ Solution 2:

$:\implies\sf \dfrac{a + b + c}{a^{-1}b^{-1} + b^{-1}c^{-1} + c^{-1}a^{-1}}\\\\\\\\:\implies\sf \dfrac{a + b + c}{\Bigg\lgroup\dfrac{1}{ab}\Bigg\rgroup + \Bigg\lgroup\dfrac{1}{bc}\Bigg\rgroup + \Bigg\lgroup\dfrac{1}{ca}\Bigg\rgroup}\\\\\\\\:\implies\sf \dfrac{a + b + c}{\Bigg\lgroup\dfrac{c + a + b}{abc}\Bigg\rgroup}\\\\\\\ :\implies\sf \dfrac{abc \Bigg\lgroup a + b + c \Bigg\rgroup}{a + b + c}\\\\\\\ :\implies \boxed{\boxed{\bf abc}}\:\bigstar\\\\\\\\ \underbrace{\pink{\bf Hence Proved}}$

$\rule{170}2$