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Express each one of the following with rational denominator:
\( \frac{b^{2}}{\sqrt{a^{2}+b^{2}}+a} \)
Given:
\( \frac{b^{2}}{\sqrt{a^{2}+b^{2}}+a} \)
To do:
We have to express the given fraction with rational denominator.
Solution:
We know that,
Rationalising factor of a fraction with denominator ${\sqrt{a}}$ is ${\sqrt{a}}$.
Rationalising factor of a fraction with denominator ${\sqrt{a}-\sqrt{b}}$ is ${\sqrt{a}+\sqrt{b}}$.
Rationalising factor of a fraction with denominator ${\sqrt{a}+\sqrt{b}}$ is ${\sqrt{a}-\sqrt{b}}$.
This implies,
Rationalising factor of the fraction with denominator $\sqrt{a^{2}+b^{2}}+a$ is $\sqrt{a^{2}+b^{2}}-a$.
Therefore,
$\frac{b^{2}}{\sqrt{a^{2}+b^{2}}+a}=\frac{b^{2}(\sqrt{a^{2}+b^{2}}-a)}{(\sqrt{a^{2}+b^{2}}+a)(\sqrt{a^{2}+b^{2}}-a)}$
$=\frac{b^{2}(\sqrt{a^{2}+b^{2}}-a)}{(\sqrt{a^{2}+b^{2}})^{2}-a^{2}}$
$=\frac{b^{2}(\sqrt{a^{2}+b^{2}}-a)}{a^{2}+b^{2}-a^{2}}$
$=\frac{b^{2}(\sqrt{a^{2}+b^{2}}-a)}{b^{2}}$
$=\sqrt{a^{2}+b^{2}}-a$
Hence, $\frac{b^{2}}{\sqrt{a^{2}+b^{2}}+a}=\sqrt{a^{2}+b^{2}}-a$.