Given that $\tan ^{-1}(2 x)+\tan ^{-1}(3 x)=\frac{\pi}{4}$
$\Rightarrow \tan ^{-1}\left(\frac{2 x+3 x}{1-2 x \times 3 x}\right)=\frac{\pi}{4}$
$\Rightarrow \frac{5 x}{1-6 x^2}=\tan \frac{\pi}{4}=1$
$\Rightarrow \quad 5 x=1-6 x^2$
$\Rightarrow \quad 6 x^2+5 x-1=0$
$\Rightarrow \quad 6 x^2+6 x-x-1=0$
$\Rightarrow \quad 6 x(x+1)-1(x+1)=0$
$\Rightarrow \quad(x+1)(6 x-1)=0$
$\Rightarrow \quad \text { Either } x+1=0 \Rightarrow x=-1 \text { which is not possible. }$
$\text { or } \quad 6 x-1=0$
$\Rightarrow \quad x=\frac{1}{6}$
