32.

Given: x = $\frac{\sqrt{{\mathrm{a}}^2 + {\mathrm{b}}^2} +\hspace{0.17em}\sqrt{{\mathrm{a}}^2 - {\mathrm{b}}^2}}{\sqrt{{\mathrm{a}}^2 + {\mathrm{b}}^2} - \sqrt{{\mathrm{a}}^2 - {\mathrm{b}}^2}}$

Use componendo and dividendo to prove that b² = $\frac{2{\mathrm{a}}^2\mathrm{x}}{{\mathrm{x}}^3 +\hspace{0.17em}1}$

Given, x = $\frac{\sqrt{{\mathrm{a}}^2 + {\mathrm{b}}^2} +\hspace{0.17em}\sqrt{{\mathrm{a}}^2 - {\mathrm{b}}^2}}{\sqrt{{\mathrm{a}}^2 + {\mathrm{b}}^2} - \sqrt{{\mathrm{a}}^2 - {\mathrm{b}}^2}}$

By componendo - dividendo,

$\frac{\mathrm{x} + 1}{\mathrm{x} - 1} = \frac{{\displaystyle \sqrt{{\mathrm{a}}^2 + {\mathrm{b}}^2} +\hspace{0.17em}\sqrt{{\mathrm{a}}^2 - {\mathrm{b}}^2} +\sqrt{{\mathrm{a}}^2 + {\mathrm{b}}^2} - \sqrt{{\mathrm{a}}^2 - {\mathrm{b}}^2}}}{{\displaystyle \sqrt{{\mathrm{a}}^2 + {\mathrm{b}}^2} +\hspace{0.17em}\sqrt{{\mathrm{a}}^2 - {\mathrm{b}}^2} - \sqrt{{\mathrm{a}}^2 + {\mathrm{b}}^2} - \sqrt{{\mathrm{a}}^2 - {\mathrm{b}}^2}}}$

$\frac{\mathrm{x} + 1}{\mathrm{x} - 1} =\frac{{\displaystyle 2\sqrt{{\mathrm{a}}^2 + {\mathrm{b}}^2}}}{2\sqrt{{\mathrm{a}}^2 - {\mathrm{b}}^2}}$

Squaring both sides,

$\frac{{\mathrm{x}}^2 +\hspace{0.17em}2\mathrm{x} +\hspace{0.17em}1}{{\mathrm{x}}^2 -\hspace{0.17em}2\mathrm{x} +\hspace{0.17em}1} = \frac{{\mathrm{a}}^2 +\hspace{0.17em}{\mathrm{b}}^2}{{\mathrm{a}}^2 - \hspace{0.17em}{\mathrm{b}}^2}$

By componendo - dividendo,

$\frac{({\mathrm{x}}^2 +\hspace{0.17em}2\mathrm{x} +\hspace{0.17em}1) +\hspace{0.17em}({\mathrm{x}}^2 -\hspace{0.17em}2\mathrm{x} +\hspace{0.17em}1)}{({\mathrm{x}}^2 +\hspace{0.17em}2\mathrm{x} +\hspace{0.17em}1) - ({\mathrm{x}}^2 -\hspace{0.17em}2\mathrm{x} +\hspace{0.17em}1)} = \frac{({\mathrm{a}}^2 +\hspace{0.17em}{\mathrm{b}}^2) + ({\mathrm{a}}^2 - \hspace{0.17em}{\mathrm{b}}^2)}{({\mathrm{a}}^2 +\hspace{0.17em}{\mathrm{b}}^2) - ({\mathrm{a}}^2 - \hspace{0.17em}{\mathrm{b}}^2)}$

$\Rightarrow \frac{2({\mathrm{x}}^2 +1)}{4\mathrm{x}} = \frac{2{\mathrm{a}}^2}{2{\mathrm{b}}^2}$

$\Rightarrow \frac{{\mathrm{x}}^2 +1}{2\mathrm{x}} = \frac{{\mathrm{a}}^2}{{\mathrm{b}}^2}$

$\Rightarrow {\mathrm{b}}^{2 }= \frac{2{\mathrm{a}}^2\mathrm{x}}{{\mathrm{x}}^2 + 1}$

Hence proved.