If $tan^{-1}2x+tan^{-1}3x=\frac{\pi}{4},$ then value of x is :

$\frac{1}{6}$

The correct answer is Option (4) → $\frac{1}{6}$

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

$⇒\tan^{-1}\frac{2x+3x}{1-6x^2}=\frac{\pi}{4}$

so $\frac{5x}{1-6x^2}⇒5x=1-6x^2$

$6x^2+5x-1=0$

$6x^2+6x-x-1=0$

$6x(x+1)-(x+1)=0$

$x=\frac{1}{6}$ as $x=-1$ doesn't satisfy original equation