Question 1205626
{{{a=-1/(1-b^2)}}}

prove that
{{{a=1/(b^2-1)}}} is same

{{{a=1/(b^2-1)}}} ...we can write {{{(b^2-1)}}} as {{{-(-b^2+1)}}}, and rearange terms inside parentheses {{{-(1-b^2)}}}

so, we have

{{{a=1/(-(1-b^2))}}} which is

{{{a=-1/(1-b^2)}}}