Question

Make b the subject of formula : $a=\frac{1+b^2}{1-b^2}$.
Hence, find $b$ when $a=1$.

Answer 1

Given,

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

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

$a{b}^2+b^2=a-1$

$b^2(a+1)=a-1$

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

$\therefore b=\sqrt{\frac{a-1}{a+1}}$

Take value of $a=1$

$\therefore b=\sqrt{\frac{1-1}{1+1}}=\frac{0}{2}=0$