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 dividendo to prove that $b^2=\frac{{2ax}^{2}x}{x^3+1}$.

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,

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

Squaring both sides :

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

$\Rightarrow (x^2+2x+1)(a^2-b^2)=(a^2+b^2)(x^2-2x+1)$

$=a^2{x}^2+2a^{2}x+a^2-b^2{x}^2-2b^{2}x-b^2$

$=a^2{x}^2-2a^{2}x+a^2+b^2{x}^2-2b^{2}x+b^2$

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

on $4a^{2}x=2b^2{x}^2+2b^2$

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

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

Hence, proved.