Question

solution of mathematics

Answer 1

please ho through the image,Hope that helps

Answer 2

Answer:

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

To Show: $x^2-y^2=\frac{-280\sqrt{3}}{121}$

First find value of x²

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

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

$x^2=\frac{25+3-10\sqrt{3}}{25+3+10\sqrt{3}}$

$x^2=\frac{28-10\sqrt{3}}{28+10\sqrt{3}}$

$x^2=\frac{14-5\sqrt{3}}{14+5\sqrt{3}}$

Now,  find value of y²

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

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

$y^2=\frac{25+3+10\sqrt{3}}{25+3-10\sqrt{3}}$

$y^2=\frac{28+10\sqrt{3}}{28-10\sqrt{3}}$

$y^2=\frac{14+5\sqrt{3}}{14-5\sqrt{3}}$

Consider,

$x^2-y^2=\frac{14-5\sqrt{3}}{14+5\sqrt{3}}-\frac{14+5\sqrt{3}}{14-5\sqrt{3}}$

$=\frac{(14-5\sqrt{3})^2-(14+5\sqrt{3})^2}{(14+5\sqrt{3})(14-5\sqrt{3})}$

$=\frac{196+75-140\sqrt{3}-196-75-140\sqrt{3}}{196-75}$

$=\frac{-280\sqrt{3}}{121}$

Hence Proved.