Question

Find the value of √3-1/√3+1 -a+b√3

Answer 1

$\huge\boxed{\texttt{\fcolorbox{Red}{aqua}{Hey Mate!!!}}}$

$<b><i><font face=Copper black size=4 color=blue>$ Here is your answer.
$\frac{ \sqrt{3} - 1}{ \sqrt{3} + 1 } = a + b \sqrt{3} \\ taking \: lhs \\ = \frac{ (\sqrt{3} - 1) }{ (\sqrt{3} + 1) } \times \frac{( \sqrt{3} - 1)}{( \sqrt{3} - 1) } \\ = \frac{ \sqrt{3}( \sqrt{3} - 1) - 1( \sqrt{3} - 1) }{( \sqrt{3}) {}^{2} - 1 {}^{2} } \\ = \frac{3 - \sqrt{3} - \sqrt{3} + 1 }{3 - 1} \\ = \frac{4 - 2 \sqrt{3} }{2} \\ = \frac{4}{2} - \frac{2 \sqrt{3} }{2} \\ = 2 - \sqrt{3} \\ = a + b \sqrt{3} \\ on \: comparing \: rhs \\ a = 2 \\ b = - 1$
Therefore the value of
a = 2
b = -1

$\large{\red{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\underline{\underline{\underline{Hope \: it \: helps \: you}}}}}}}}}}}}}}}$

Answer 2

Here is your answer
$\frac{ \sqrt{3} - 1}{ \sqrt{3} + 1 } \times \frac{ \sqrt{3} - 1}{ \sqrt{3} - 1} \\ = \frac{3 - \sqrt{3} - \sqrt{3} + 1}{ \sqrt{3} {}^{2} - 1 {}^{2} } \\ = \frac{4 - 2 \sqrt{3} }{3 - 1} \\ = 2 - \sqrt{3} \\ = a + b \sqrt{3}$
so
a = 2
b= -1

Hope it helps