Past exam of the mathematics course of the University of Cambridge / 2024 / ia / Paper 2 / 2A (Differential Equations)

Set $x=y=0$ in the functional equation:The denominator is defined. If $f(0)\ne 0$, cancellation would give $f(0{)}^2=-1$, impossible for a real-valued function. Hence

Using the addition law with $y=h$,Differentiability makes $f$ continuous, so $f(h)\to f(0)=0$, andWriting $C=f^{\prime }(0)$ and taking the limit gives

Sinceintegration givesPart (i) gives $D=0$, so the tangent addition functional equation has the solutionsfor constants $C$ such that no pole lies in $(-a,a)$. Conversely, the tangent addition formula verifies the functional equation. If $f^{\prime }(0)=2$, then $C=2$, andon any such interval, necessarily with $a\le \pi /4$.