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.