Mark the tick against the correct answer in the following:

Given: $\tan ^{-1} x+\tan ^{-1} 3=\tan ^{-1} 8$

To Find: The value of $x$

Here $\tan ^{-1} x+\tan ^{-1} 3=\tan ^{-1} 8$ can be written as

$\tan ^{-1} x=\tan ^{-1} 8-\tan ^{-1} 3$

Since we know that $\tan ^{-1} x$ - $\tan ^{-1} y=\tan ^{-1}\left(\frac{x-y}{1+x y}\right)$

$\tan ^{-1} x=\tan ^{-1} 8-\tan ^{-1} 3=\tan ^{-1}\left(\frac{8-3}{1+(8 \times 3)}\right)$

$=\tan ^{-1}\left(\frac{5}{25}\right)$

$=\tan ^{-1}\left(\frac{1}{5}\right)$

$\Rightarrow x=\frac{1}{5}$