Q.Solve for $x$: $\tan^{-1}(2x)+\tan^{-1}(3x)=\dfrac{\pi}{4}$.

The formula $\tan^{-1}x+\tan^{-1}y=\tan^{-1}\!\left(\frac{x+y}{1-xy}\right)$ follows from the tangent addition formula, but only holds directly when $xy<1$, which keeps $\tan^{-1}x+\tan^{-1}y$ inside the principal branch $(-\pi/2,\pi/2)$; when $x,y>0$ and $xy>1$, the true sum exceeds $\pi/2$ and a correction of $+\pi$ must be added instead: $\tan^{-1}x+\tan^{-1}y=\pi+\tan^{-1}\!\left(\frac{x+y}{1-xy}\right)$. Setting $y=x$ gives the double-angle form $2\tan^{-1}x=\tan^{-1}!\ …