We have $\tan^{-1}\large\frac{1}{a-1}$$=\tan^{-1}\large\frac{1}{x}$$+\tan^{-1}\large\frac{1}{a^2-x+1}$
$\Rightarrow \tan^{-1}\large\frac{1}{a-1}$$=\tan^{-1}\large\frac{1/x+1/a^2-x+1}{1-1/x(a^2-x+1)}$
$\Rightarrow \large\frac{1}{a-1}=\frac{a^2+1}{(a^2+1)(x-x^2-1)}$
$x^2-x(a^2+1)+1+(a-1)(a^2+1)=0$
$x^2-a^2-(a^2+1)(x-a)=0$
$(x-a)(x+a-a^2-1)=0$
$x=a$
Or $x=a^2-a+1$
Hence (c) is the correct option.