Approximate a shifted tangent using an exact identity

Both copies print only the answer. Independently put u=x/3 and use

$\tan(\pi/4+u)=\frac{1+\tan u}{1-\tan u}.$

Since $\tan u=u+O(u^3)$, its cubic correction cannot affect the requested quadratic coefficients near u=0.

Through the quadratic order,

$\begin{aligned}&\frac{1+u}{1-u}\\ & =(1+u)(1+u+u^2+\cdots)\\ & =1+2u+2u^2+\cdots.\end{aligned}$

Substitute u=x/3 to get the printed answer $1+\tfrac23x+\tfrac29x^2$. This is an approximation to the original shifted tangent, not an exact finite identity.