Question

Rationalise the denominator:-
$\sqrt{3} - 1 \div \sqrt{3} + 1$

Answer 1

$\huge\red{Hi\: Friend}$

Here is your answer
$= \frac{ \sqrt{3} - 1 }{ \sqrt{3} + 1 } \times \frac{ \sqrt{3} - 1}{ \sqrt{3} - 1} \\ = \frac{ \sqrt{3}( \sqrt{3} - 1) - 1( \sqrt{3} - 1)}{( \sqrt{3}) {}^{2} - 1 {}^{2} } \\ = \frac{3 - \sqrt{3} - \sqrt{3} + 1}{3 - 1} \\ = \frac{4- 2\sqrt{3} }{2} \\ = \frac{4}{2} - \frac{2 \sqrt{3} }{2} \\ = 2 - \sqrt{3}$
$<marquee>Hope it helps you$

Answer 2

$\large\mathsf{Ahoy!!}$
$<i>$
Given $\rightarrow$

$\frac{ \sqrt{3} - 1}{ \sqrt{3} + 1 } \\ \\ = \frac{ \sqrt{3} - 1 }{ \sqrt{3} + 1} \times \frac{ \sqrt{3} - 1}{ \sqrt{3} - 1} \\ \\ = \frac{( \sqrt{3} - 1)( \sqrt{3} - 1) }{ (\sqrt{3} + 1)( \sqrt{3} - 1) }$

{ Applying identity :

( a + b ) ( a - b ) = a² - b² }

$= \frac{ {( \sqrt{3} - 1)}^{2} }{ ( { \sqrt{3}) }^{2} - {(1)}^{2} }$

{ Applying identity :

( a - b )² = a² + b² - 2ab }

$= \frac{ ({ \sqrt{3}) }^{2} + {1}^{2} - 2( \sqrt{3} )(1)}{3 - 1} \\ \\ = \frac{3 + 1 - 2 \sqrt{3} }{2} \\ \\ = \frac{4 - 2 \sqrt{3} }{2} \\ \\ = \frac{2 \: (2 - \sqrt{3}) }{2} \\ \\ = 2 - \sqrt{3}$

$\implies$TooBusy ¯\_(ツ)_/¯

$\implies$BeBrainly