Question

Given $x=\frac{\sqrt{a^2+b^2}+\sqrt{a^2-b^2}}{\sqrt{a^2+b^2}-\sqrt{a^2-b^2}}$. Using componendo and dividendo, prove that $b^2=\frac{2a^{2}x}{x^2+1}$.

Answer 1

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

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

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

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

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

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

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

$\Rightarrow 4x{a}^2=2x^2{b}^2+2b^2$

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

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

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