The solution of \(\tan^{-1}2x+\tan^{-1}3x=\frac{\pi}{4}\) is

\(1\) \(\frac{1}{3}\) \(\frac{1}{6}\) \(-1\)

\(\frac{1}{6}\)

\(\tan^{-1}2x+\tan^{-1}3x=\frac{\pi}{4}\) $⇒\tan^{-1}(\frac{2x+3x}{1-6x^2})=\frac{\pi}{4}$,   $6x^2<1$ $5x=1-6x^2$ $⇒6x^2+5x-1=0$, $6x^2<1$ $6x^5+6x-x-1=0$ so $x=\frac{1}{6},-1$ $x=\frac{1}{6}$