Q.Prove that $\tan^{-1}1+\tan^{-1}2+\tan^{-1}3=\pi$.

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 $2\tan^{-1}x=\tan^{-1}\!\left(\frac{2x}{1-x^2}\right)$ for $|x|<1$, with companion forms $2\tan^{-1}x=\sin^{-1}\!\left(\frac{2x}{1+x^2}\right)$ for $|x|\le1$ and $2\tan^{-1}x=\cos^{-1}\!\left(\frac{1-x^2}{1+x^2}\right)$ for $x\ge0$.