Mark the tick against the correct answer in the following:

Given: $\tan ^{-1} 3 x+\tan ^{-1} 2 x=\frac{\pi}{4}$

To Find: The value of $x$

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

$\Rightarrow \tan ^{-1} 3 x+\tan ^{-1} 2 x=\tan ^{-1}\left(\frac{3 x+2 x}{1-(3 x \times 2 x)}\right)$

$=\tan ^{-1}\left(\frac{5 x}{1-6 x^{2}}\right)$

Now since $\tan ^{-1} 3 x+\tan ^{-1} 2 x=\frac{\pi}{4}$

$\tan ^{-1} 3 x+\tan ^{-1} 2 x=\tan ^{-1} 1\left(\cdots \tan \frac{\pi}{4}=1\right)$

$\Rightarrow \tan ^{-1}\left(\frac{5 x}{1-6 x^{2}}\right)=\tan ^{-1} 1$

$\Rightarrow \frac{5 x}{1-6 x^{2}}=1$

$\Rightarrow 6 x^{2}+5 x-1=0$

$\Rightarrow x=\frac{1}{6}$ or $x=-1$