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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.

For S-wave scattering, write the radial wavefunction as $R_0(r)=u(r)/r$. In the attractive shell $a<r\le 2a$, the hard-core boundary condition $u(a)=0$ givesOutside the potential, chooseContinuity of the logarithmic derivative at $r=2a$ yieldsPut $C=\kappa \cot (\kappa a)$. The tangent addition formula givesand solving for $\tan {\delta }_0$ gives

Let $\Delta =\Omega -{\varphi }_d$. From part (iii), $\varphi -{\varphi }_d=\Delta +(\varphi -\Omega )$ and $\tan (\varphi -\Omega )=\cos I\tan f$. The tangent addition formula therefore givesor, without singular coordinate tangents,For $I=\pi /2-I^{\prime }$ with $I^{\prime }\ll 1$, smooth choice of the angular branch gives

The tangent addition formula givesHence $\boxed{f(v)=\tan v}$. The restriction $xy\ne 1$ excludes the poles of tangent that can arise in the possible range $-\pi <v<\pi$. One should not replace this identity by $v=\arctan u$ globally: the principal inverse tangent can differ from $v$ by an integer multiple of $\pi$.