If $\tan ^{-1} 3+\tan ^{-1} x=\tan ^{-1} 8$, then $x=$

Solution:
(b) $\frac{1}{5}$
We know that $\tan ^{-1} x+\tan ^{-1} y=\tan ^{-1} \frac{x+y}{1-x y}$.
Now,
$\tan ^{-1} 3+\tan ^{-1} x=\tan ^{-1} 8$
$\Rightarrow \tan ^{-1}\left(\frac{3+x}{1-3 x}\right)=\tan ^{-1} 8$
$\Rightarrow \frac{3+x}{1-3 x}=8$
$\Rightarrow 3+x=8-24 x$
$\Rightarrow 3-8=-24 x-x$
$\Rightarrow-5=-25 x$
$\Rightarrow x=\frac{5}{25}=\frac{1}{5}$