Question

Solve for x in the following equation

${tan}^{-1}\frac{1}{a-1}={tan}^{-1}\frac{\left(1/x\right)+1/(a^2-x+1)}{1-1/\left[x\right(a^2-x+1\left)\right]}$

Answer 1

${tan}^{-1}\frac{1}{a-1}={tan}^{-1}\frac{\left(1/x\right)+1/(a^2-x+1)}{1-1/\left[x\right(a^2-x+1\left)\right]}$
$\therefore$ $\frac{1}{a-1}=\frac{a^2+1}{(a^2+1).x-x^2-1}$
$\therefore$ $x^2-x(a^2+1)+1+(a-1)(a^2+1)=0$.
or $(x^2-a^2)-(a^2+1)(x-a)=0$
or $(x-a)(x+a-a^2-1)=0$
$\therefore$ $x=a$ or $a^2-a+1$