We can define $\sin$ and $\cos$ as the solutions to the DE $$f''(x) = -f(x)$$ subject to certain initial conditions.
The above DE has constant coefficients, but I'd like to use a similar approach to defining $\tan$, and I'm happy generalize to rational functions, for example $$f''(x) = x^3 f'(x) -\frac{3x}{1-x^2} f(x)$$ would be a valid candidate.
Anyway:
Question. Can $\mathrm{tan}$ be defined by an IVP with rational functions as coefficients?
Note that $$f'(x) = \frac{1}{1+x^2}$$ is solved by $f(x) = \mathrm{arctan}(x),$ so that's a near miss.
No. Functions which can be defined in this way are said to be holonomic, and $\tan x$ is known not to be holonomic. This is because holonomic functions have finitely many singularities (the singularities of the denominators of the rational functions involved), while $\tan x$ has infinitely many singularities.