$\begin{aligned} & \tan ^{-1} x+\tan ^{-1} 2 x=\frac{\pi}{4} ; x>0 \\ & \Rightarrow \tan ^{-1} 2 x=\frac{\pi}{4}-\tan ^{-1} x \end{aligned}$ Taking tan both sides $\begin{aligned} & \Rightarrow 2 \mathrm{x}=\frac{1-\mathrm{x}}{1+\mathrm{x}} \\ & \Rightarrow 2 \mathrm{x}^2+3 \mathrm{x}-1=0 \\ & \mathrm{x}=\frac{-3 \pm \sqrt{9+8}}{8}=\frac{-3 \pm \sqrt{17}}{8} \end{aligned}$ Only possible $x=\frac{-3+\sqrt{17}}{8}$