Question

x=1/2[√(a/b) -√(b/a)]. then prove that [2a√(1+x^2)]/[x+√(1+x^2)]=a+b​

Answer 1

Given :

$\\ \bullet \tt \: x = \frac{1}{2} \bigg( \sqrt{ \frac{a}{b} } - \sqrt{ \frac{b}{a} } \bigg)$

To Prove that :

$\\ \bullet \tt \: \frac{2a( \sqrt{1 + {x}^{2} }) }{(x + \sqrt{1 + {x}^{2}) } } = a + b$

Proof :

•Given,

$\\ \bullet \sf \: x = \frac{1}{2} \bigg( \sqrt{ \frac{a}{b} } - \sqrt{ \frac{b}{a} } \bigg)$

$\\ \bullet\bf{Let , \: t=\dfrac{b}{a}}$

•(since square root is defined on t, t cannot be negative)

then,

$\bullet \boxed{ \sf{x = \frac{1}{2} \bigg( \sqrt{ \frac{1}{t} } - \sqrt{t} \bigg)}}$

$\\ \implies \sf \: x = \frac{1}{2} \bigg( \ \sqrt{ \frac{1}{t} } - \sqrt{t} \bigg) \\ \\ \\ \implies \sf \: x = \frac{1}{2} \bigg( \frac{1 - t}{ \sqrt{t} } \bigg) \\ \\ \\ \implies \sf \: 2x \sqrt{t} = 1 - t$

$\\ \bullet\red{\large\underline{\sf{Squaring \: both \: sides, \: we \: get :}}}$

$\\ \implies \sf \: 4 {x}^{2} t = 1 + {t}^{2} - 2t \\ \\ \\ \implies \sf \: {t}^{2} - 2t(1 + 2 {x}^{2} ) + 1 = 0 \\ \\ \\ \implies \sf \: t = \frac{2(1 + 2 {x}^{2} ) \pm \: \sqrt{4} {(1 + 2 {x}^{2} )}^{2} - 4}{2}$

But, t > 0

$\\ \implies \sf \: t = (1 + 2 {x}^{2} ) + \sqrt{ {(1 + 2 {x}^{2} )}^{2} - 1 } \\ \\ \\ \implies \sf \:t = (1 + 2 {x}^{2} ) + \sqrt{(1 + 4 {x}^{2} + 4 {x}^{4} - 1} \\ \\ \\ \implies \sf \: t = (1 + 2 {x}^{2} ) + \sqrt{4 {x}^{2} + 4 {x}^{4} } \\ \\ \\ \implies \sf \: t = (1 + 2 {x}^{2} ) + 2x \sqrt{ {x}^{2} + 1 } \\ \\ \\ \implies \sf \: t + 1 = 1 + (1 + 2 {x}^{2} ) + 2x \sqrt{ {x}^{2} + 1} \\ \\ \\ \implies \sf \: t + 1 = 2(1 + {x}^{2} + x \sqrt{ {x}^{2} + 1 } )$

But t=b/a , so the above expressions becomes :

$\\ \implies \sf \: \frac{b}{a} + 1 = 2(1 + {x}^{2} + x \sqrt{ {x}^{2} + 1 } ) \\ \\ \\ \implies \sf \: \bigg( \frac{a + b}{a} \bigg) = 2(1 + {x}^{2} + x \sqrt{ {x}^{2} + 1} ) \\ \\ \\ \implies \sf \: a + b = 2a(1 + {x}^{2} + x \sqrt{ {x}^{2} + 1 } )$

$\\ \bullet\green{\large\underline{\sf{Taking \: \sqrt{x^{2}+1} \: as \: a \: common ,\: we \: get :}}}$

$\\ \implies \sf \: a + b = 2a \sqrt{ {x}^{2} + 1} (x + \sqrt{ {x}^{2} + 1 } )$

Hence,

$\pink \bigstar \boxed{ \sf{2a \sqrt{ {x}^{2} + 1 } (x + \sqrt{ {x}^{2} + 1} )= a + b}}$