Question

Given $x=\frac{\sqrt{a^2+b^2}+\sqrt{a^2-b^2}}{\sqrt{a^2+b^2}-\sqrt{a^2-b^2}}$
use componendo and dividend0 to prove that :
$b^2=\frac{2a^{2}x}{x^2+1}$.

Answer 1

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

Applying Componendo and Dividendo

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

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

Squaring both sides

${\left(\frac{x+1}{x-1}\right)}^2=\frac{a^2+b^2}{a^2-b^2}$

Applying Componendo & Dividendo

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

$\frac{2(x^2+1)}{2\left(2x\right)}=\frac{a^2}{b^2}$

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

$b^2=\frac{2a^{2}x}{x^2+1}$ (RHS)

(Proved)