Question

anyone solve this question please of rs agarwal,??? ​

Answer 1

Step-by-step explanation:

Given,

$x = \frac{5 - \sqrt{3} }{5 + \sqrt{3} } \: \: and \: \: y = \frac{5 + \sqrt{3} }{5 - \sqrt{3} }$

Now,

${x}^{2} - {y}^{2} = { (\frac{5 - \sqrt{3} }{5 + \sqrt{3} } )}^{2} - { (\frac{5 + \sqrt{3} }{5 - \sqrt{3} } )}^{2}$

$= > {x}^{2} - {y}^{2} = ( \frac{5 - \sqrt{3} }{5 + \sqrt{3} } + \frac{5 + \sqrt{3} }{5 - \sqrt{3} } )( \frac{5 - \sqrt{3} }{5 + \sqrt{3}} - \frac{5 + \sqrt{3} }{5 - \sqrt{3} } )$

$= > {x}^{2} - {y}^{2} = ( \frac{ {(5 - \sqrt{3} )}^{2} + {(5 + \sqrt{3} )}^{2} }{(5 + \sqrt{3})(5 - \sqrt{3}) } )( \frac{ {(5 - \sqrt{3} )}^{2} - {(5 + \sqrt{3} )}^{2} }{(5 + \sqrt{3})(5 - \sqrt{3}) } )$

$= > {x}^{2} - {y}^{2} = ( \frac{2(25 + 3)}{(25 - 3)})( \frac{ - 4.5. \sqrt{3} }{(25 - 3)})$

$= > {x}^{2} - {y}^{2} = \frac{ - 56 \times 20 \sqrt{3} }{22 \times 22}$

Answer 2

Answer:

Now thank my answers

otherwise I will report your answers