If $x=\frac{\sqrt{a^{2}+b^{2}}+\sqrt{a^{2}-b^{2}}}{\sqrt{a^{2}+b^{2}}-\sqrt{a^{2}-b^{2}}}$, then prove that $b^{2} x^{2}-2 a^{2} x+b^{2}=0$.

Given:

$x=\frac{\sqrt{a^{2}+b^{2}}+\sqrt{a^{2}-b^{2}}}{\sqrt{a^{2}+b^{2}}-\sqrt{a^{2}-b^{2}}}$

To do:

We have to prove that $b^{2} x^{2}-2 a^{2} x+b^{2}=0$.
Solution:
$\begin{array}{l} x=\frac{\sqrt{a^{2} +b^{2}} +\sqrt{a^{2} -b^{2}}}{\sqrt{a^{2} +b^{2}} -\sqrt{a^{2} -b^{2}}}\\ x=\frac{\left(\sqrt{a^{2} +b^{2}} +\sqrt{a^{2} -b^{2}}\right)\left(\sqrt{a^{2} +b^{2}} +\sqrt{a^{2} -b^{2}}\right)}{\left(\sqrt{a^{2} +b^{2}} -\sqrt{a^{2} -b^{2}}\right)\left(\sqrt{a^{2} +b^{2}} +\sqrt{a^{2} -b^{2}}\right)}\\ x=\frac{\left(\sqrt{a^{2} +b^{2}}\right)^{2} +\left(\sqrt{a^{2} -b^{2}}\right)^{2} +2\sqrt{a^{2} +b^{2}} \times \sqrt{a^{2} -b^{2}}}{\left(\sqrt{a^{2} +b^{2}}\right)^{2} -\left(\sqrt{a^{2} -b^{2}}\right)^{2}}\\ x=\frac{a^{2} +b^{2} +a^{2} -b^{2} +2\sqrt{a^{4} -b^{4}}}{a^{2} +b^{2} -\left( a^{2} -b^{2}\right)}\\ x=\frac{2a^{2} +2\sqrt{a^{4} -b^{4}}}{2b^{2}}\\ x=\frac{a^{2} +\sqrt{a^{4} -b^{4}}}{b^{2}}\\ Therefore,\\ b^{2} x^{2} -2a^{2} x+b^{2} =b^{2}\left(\frac{a^{2} +\sqrt{a^{4} -b^{4}}}{b^{2}}\right)^{2} -2a^{2}\left(\frac{a^{2} +\sqrt{a^{4} -b^{4}}}{b^{2}}\right) +b^{2}\\ =\frac{a^{4} +a^{4} -b^{4} +2a^{2}\sqrt{a^{4} -b^{4}} -2a^{4} -2a^{2}\sqrt{a^{4} -b^{4}} +b^{4}}{b^{2}}\\ =\frac{2a^{4} -2a^{4}}{b^{2}}\\ =0 \end{array}$

Hence proved.

Tutorialspoint

Simply Easy Learning