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

Answer 1

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

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

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

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

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