if

We know that $\tan ^{-1} x+\cot ^{-1} x=\frac{\pi}{2}$.

We have

$\tan ^{-1}(\sqrt{3})+\cot ^{-1} x=\frac{\pi}{2}$

$\Rightarrow \tan ^{-1}(\sqrt{3})=\frac{\pi}{2}-\cot ^{-1} x$

$\Rightarrow \tan ^{-1}(\sqrt{3})=\tan ^{-1} x$

$\Rightarrow x=\sqrt{3}$

$\therefore x=\sqrt{3}$