Rationalize the denominator of the following expression: $\frac{n^{2}}{\sqrt{m^{2}+n^{2}}+m}$

To rationalize the denominator,multiply the numerator and the denominator by the conjugate of the denominator,which is $\sqrt{m^{2}+n^{2}}-m$.
$\frac{n^{2}}{\sqrt{m^{2}+n^{2}}+m} = \frac{n^{2}}{\sqrt{m^{2}+n^{2}}+m} \times \frac{\sqrt{m^{2}+n^{2}}-m}{\sqrt{m^{2}+n^{2}}-m}$
Using the identity $(a+b)(a-b) = a^{2}-b^{2}$ in the denominator:
$= \frac{n^{2}(\sqrt{m^{2}+n^{2}}-m)}{(\sqrt{m^{2}+n^{2}})^{2} - (m)^{2}}$
$= \frac{n^{2}(\sqrt{m^{2}+n^{2}}-m)}{m^{2}+n^{2}-m^{2}}$
$= \frac{n^{2}(\sqrt{m^{2}+n^{2}}-m)}{n^{2}}$
$= \sqrt{m^{2}+n^{2}}-m$