If $x=\frac{5-\sqrt{3}}{5+\sqrt{3}}$ and $y=\frac{5+\sqrt{3}}{5-\sqrt{3}}$, show that $x^{2}-y^{2}=-\frac{10 \sqrt{3}}{11}$
Disclaimer: The question is incorrect.
$x=\frac{5-\sqrt{3}}{5+\sqrt{3}}$
$\Rightarrow x=\frac{5-\sqrt{3}}{5+\sqrt{3}} \times \frac{5-\sqrt{3}}{5-\sqrt{3}}$
$\Rightarrow x=\frac{(5-\sqrt{3})^{2}}{5^{2}-(\sqrt{3})^{2}}$
$\Rightarrow x=\frac{25+3-10 \sqrt{3}}{25-3}$
$\Rightarrow x=\frac{28-10 \sqrt{3}}{22}$
$\Rightarrow x=\frac{14-5 \sqrt{3}}{11}$
$y=\frac{5+\sqrt{3}}{5-\sqrt{3}}$
$\Rightarrow y=\frac{5+\sqrt{3}}{5-\sqrt{3}} \times \frac{5+\sqrt{3}}{5+\sqrt{3}}$
$\Rightarrow y=\frac{(5+\sqrt{3})^{2}}{5^{2}-(\sqrt{3})^{2}}$
$\Rightarrow y=\frac{25+3+10 \sqrt{3}}{25-3}$
$\Rightarrow y=\frac{28+10 \sqrt{3}}{22}$
$\Rightarrow y=\frac{14+5 \sqrt{3}}{11}$
$\therefore x^{2}-y^{2}$
$=\left(\frac{14-5 \sqrt{3}}{11}\right)^{2}-\left(\frac{14+5 \sqrt{3}}{11}\right)^{2}$
$=\frac{196+75-140 \sqrt{3}}{121}-\frac{196+75+140 \sqrt{3}}{121}$
$=\frac{271-140 \sqrt{3}}{121}-\frac{271+140 \sqrt{3}}{121}$
$=\frac{271-140 \sqrt{3}-271-140 \sqrt{3}}{121}$
$=\frac{-280 \sqrt{3}}{121}$
The question is incorrect. Kindly check the question.
The question should have been to show that $x-y=-\frac{10 \sqrt{3}}{11}$.
$\therefore x-y$
$=\frac{14-5 \sqrt{3}}{11}-\frac{14+5 \sqrt{3}}{11}$
$=\frac{14-5 \sqrt{3}-14-5 \sqrt{3}}{11}$
$=\frac{-10 \sqrt{3}}{11}$
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.