By componendo-dividendo on $\dfrac{1+bx}{1-bx} = \dfrac{(1+ax)^2}{(1-ax)^2}$, the next simplified form is—

Question

Accepted Answer

$\dfrac{2}{2bx} = \dfrac{2(1 + a^2x^2)}{4ax}$

Answer

$\dfrac{1}{bx} = ax$

Answer

$\dfrac{1+bx}{1+ax} = 1$

Answer

$bx = ax$

By componendo-dividendo on 1+bx1−bx=(1+ax)2(1−ax)2\dfrac{1+bx}{1-bx} = \dfrac{(1+ax)^2}{(1-ax)^2}1−bx1+bx​=(1−ax)2(1+ax)2​, the next simplified form is—