If you're truly interested in why FullSimplify[] fails, I can offer some hints. I don't know for sure, in fact, but its occasional failures like this seem understandable to me, and I can share my understanding.

FullSimplify[] has an extensive set of transformations. But it's finite. I'm not sure what it is in it and what is not. Somewhat relevant here is that it has a general addition/subtraction identity for tangent for a general angle a but not for the specific angle Pi/4:

(1 - Tan[t])/(1 + Tan[t]) // FullSimplify
(*  -1 + 2/(1 + Tan[t])  *)

(Tan[a] - Tan[t])/(1 + Tan[a] Tan[t]) // FullSimplify
(*  Tan[a - t]  *)

FullSimplify[] is a minimizer. I don't think the above is the snag that stops it. FullSimplify[] will take a step if a transformation results in an expression that is less complex. If it is more complex, it may try a couple of more steps to see if it gets over a hump to a less complex expression. If it takes several complicated steps before one gets to the desired form, FullSimplify[] may stop before it gets there, even if the required transformations are in its toolbox. I think this is probably what happens.

FullSimplify has plug-in capabilities to enhance its performance. See the options TransformationFunctions and ComplexityFunction.

Examples

Some example for you. Sorry I don't have time to explain.

Series approach

(* match expression to c Tan[k t + ArcTan[a]] == c(a+Tan[k t])/(1-a Tan[k t]) via Series *)
tanSumReduce // ClearAll;
tanSumReduce[expr : _. _Tan] := expr; (* skip k*Tan[x] *)
tanSumReduce[expr_] := With[{arg = Cases[expr, Tan[x_] :> x, Infinity]},
   Module[{c, a, k, paramsol, t},
     paramsol = SolveAlways[
       Normal@Series[
          c (a + Tan[k t])/(1 - a Tan[k t]) - 
           (expr /. First@arg -> t), {t, 0, 3}] == 0, t];
     Replace[
      expr,
      {e_ /; Length@paramsol > 0 :> 
        With[{res = c Tan[k*First@arg + ArcTan@a] /. Last@paramsol},
         res /; Simplify[res - expr] === 0]}
      ]
     ] /; Length[arg] > 0];

Assuming[2 Pi < dtheta < 3 Pi,
 FullSimplify[ArcTan[(1 - Sin[dtheta/2])/Cos[dtheta/2]],
  TransformationFunctions -> {Automatic, tanSumReduce}]
 ]
(*  1/4 (-dtheta + \[Pi])  *)

Differential equations approach

One needs to specify the independent variable. It is not deduced automatically.

(* match c Tan[k t + t0] via D[y'[k t]/c/k==(y[k t]/c)^2+1, t] *)
tanReduce // ClearAll;
tanReduce[v_][expr_] := Module[{y, v0, amp, freq, sign},
   y[Pattern[Evaluate[v], _]] = expr;
   amp = Sqrt@Simplify[y[v] (2 y'[v]^2 - y''[v] y[v])/y''[v]];
   v0 = ArcTan[y[0]/amp];
   freq = 
    Sqrt@Simplify[y''[v] (2 y'[v]^2 - y''[v] y[v])/(4 y[v] y'[v]^2)];
   sign = Sign[Limit[y'[v], v -> (v0)/freq]];
   amp*Tan[sign*freq*v + v0] /; 
    Simplify[y[v] - amp*Tan[sign*freq*v + v0]] === 0
   ];

Simplify[ArcTan[(1 - Sin[dtheta/2])/Cos[dtheta/2]], 0 < dtheta < Pi, 
 TransformationFunctions -> {Automatic, tanReduce[dtheta]}]
(*  1/4 (-dtheta + \[Pi])  *)

Similar example, see here

Regards M.I.

That's a nice example, thank you, and one of the things they did (and I did) was a series expansion, which gives the equivalent simple linear expression with higher order terms zero (at least as far as one wants to set the order!). They also evaluated the derivative and showed how simple it was - neat trick.

Next stop - let's see how grok3 AI does. (Edited). grok3 solved it after a number of tries and a few screenfuls of test functions, It found that

( 1-Sin[A] ) / Cos[A] is the same as Tan[Pi/4 -A/2] and used that.