We have $\tan^{-1}4\text{x}+\tan^{-1}6\text{x}=\frac{\pi}{4}$
$\Rightarrow\tan(\tan^{-1}4\text{x}+\tan^{-1}6\text{x})=\tan\frac{\text{x}}4{}$
$\Rightarrow\frac{\tan(\tan^{-1}4\text{x})+\tan(\tan^{-1}6\text{x})}{1-\tan(\tan^{-1}4\text{x}).\tan(\tan^{-1}6\text{x})}=1$
$\Rightarrow\frac{4\text{x}+6\text{x}}{1-4\text{x}.6\text{x}}=1$
$\Rightarrow\frac{10\text{x}}{1-24\text{x}^2}=1$
$\Rightarrow24\text{x}^2+10\text{x}-1=0$
$\Rightarrow24\text{x}^2+12\text{x}-2\text{x}-1=0$
$\Rightarrow12\text{x}(2\text{x}+1)-1(2\text{x}+1)=0$
$\Rightarrow(2\text{x}+1)(12\text{x}-1)=0$
$\Rightarrow\text{x}=-\frac{1}{2},\frac{1}{12}$
But $\text{x}=\frac{-1}{2}$ does not satisfy the equation as the LHS will become negative
Therefore, the value of x is $\frac{1}{12}.$
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