Arctangent addition formula ($\arctan A+\arctan B=\arctan ((A+B)/(1-AB))(AB<1)$)

For real $A,B$ with $AB<1$, the principal inverse tangent satisfiesTo prove it, multiply the complex numbers $1+iA$ and $1+iB$. Their product is $(1-AB)+i(A+B)$, whose positive real part places its complex argument in $(-\pi /2,\pi /2)$. The individual complex arguments are $\arctan A$ and $\arctan B$, and their sum is in $(-\pi ,\pi )$. Multiplication adds these complex arguments, so the positive real part forces the sum into the same principal range and gives the displayed formula.

If $AB>1$, both $A,B$ have the same sign. Add $\pi$ to the right-hand side when $A,B>0$, and subtract $\pi$ when $A,B<0$. If $AB=1$, the sum is $\pi /2$ or $-\pi /2$ according to the sign of $A+B$. These cases record the branch information that can be lost by applying the tangent addition formula alone.