How do you make Tan[A+B] expansion give the well known high school answer?

Question

What command makes Tan[A+B] expand to (Tan[A]+Tan[B])/(1-Tan[A]Tan[B])? TrigExpand does not give it directly.

Answer 1

It is in the Wolfram Knowledgebase

Clear["Global`*"]
Entity["MathematicalFunction", "Tan"]["AdditionFormulas"][[1]][
  A, B] // Activate
(* Tan[A + B] == (Tan[A] + Tan[B])/(1 - Tan[A] Tan[B]) *)

Verifying,

% // Simplify
(* True *)

EDIT: Converting the equality into a rule

rule = Tan[A_ + B_] :> Evaluate[%%[[-1]]]
(* Tan[A_ + B_] :> (Tan[A] + Tan[B])/(1 - Tan[A] Tan[B]) *)

Handling multiple addends in argument to Tan,

Simplify[Tan[a + b + c + d] //. rule, ExcludedForms -> _Tan]
(* (Tan[b] + Tan[c] + Tan[d] - Tan[b] Tan[c] Tan[d] - 
   Tan[a] (-1 + Tan[c] Tan[d] + Tan[b] (Tan[c] + Tan[d])))/(1 - 
   Tan[c] Tan[d] - Tan[b] (Tan[c] + Tan[d]) + 
   Tan[a] (-Tan[c] - Tan[d] + Tan[b] (-1 + Tan[c] Tan[d]))) *)

Verifying,

% // Simplify
(* Tan[a + b + c + d] *)

Answer 2

I think the easiest way is the following:

tansum = Tan[a_ + b_] :> (Tan[a] + Tan[b])/(1 - Tan[a] Tan[b]); 

Then we run

Tan[A + B] /. tansum

(Tan[A] + Tan[B])/(1 - Tan[A] Tan[B])

Likewise for other trigonometric identities

Edit: after the comment for generalization for $n$-angles.

If we try to do it like this for the case $n=3$ it seems to be working

Tan[a1 + a2 + a3] /. tansum /. tansum // Factor

(-Tan[a1] - Tan[a2] - Tan[a3] + Tan[a1] Tan[a2] Tan[a3])/(-1 + Tan[a1] Tan[a2] + Tan[a1] Tan[a3] + Tan[a2] Tan[a3])

Answer 3

You can do a step-by-step derivation, suppressing the evaluation using a change of case.

expr0 = Tan[a + b]

$$\tan (a+b)$$

expr1 = TrigExpand[expr0] /. {Cos -> cos, Sin -> sin}

$$\frac{\sin (a) \cos (b)}{\cos (a) \cos (b)-\sin (a) \sin (b)}+\frac{\cos (a) \sin (b)}{\cos (a) \cos (b)-\sin (a) \sin (b)}$$

expr2 = expr1 /. {sin[a_] -> tan[a] cos[a] }

$$\frac{\cos (a) \tan (a) \cos (b)}{\cos (a) \cos (b)-\cos (a) \tan (a) \cos (b) \tan (b)}+\frac{\cos (a) \cos (b) \tan (b)}{\cos (a) \cos (b)-\cos (a) \tan (a) \cos (b) \tan (b)}$$

expr3 = Factor[expr2] /. tan -> Tan

$$-\frac{\tan (a)+\tan (b)}{\tan (a) \tan (b)-1}$$

expr3 // Simplify

Tan[a+b]

---

The same procedure for Tan[a+b+c+d] results in:

$$-\frac{\tan (a) \tan (b) \tan (c)+\tan (a) \tan (b) \tan (d)+\tan (a) \tan (c) \tan (d)-\tan (a)+\tan (b) \tan (c) \tan (d)-\tan (b)-\tan (c)-\tan (d)}{\tan (a) \tan (b) \tan (c) \tan (d)-\tan (a) \tan (b)-\tan (a) \tan (c)-\tan (a) \tan (d)-\tan (b) \tan (c)-\tan (b) \tan (d)-\tan (c) \tan (d)+1}$$

Answer 4

If the question concerns the derivation, not the application, of the "high school formula" one can derive it , similar to @Syed's answer, in a direct way :

((Tan[a + b] // TrigExpand) /. {a ->  ArcTan[ua], b ->  ArcTan[ub]} //TrigExpand // Simplify) /. {ua -> Tan[a ], ub -> Tan[b ]}
(*(Tan[a] + Tan[b])/(1 - Tan[a] Tan[b])*)

Answer 5

Lots of good answers, but here's a trick I think is worth knowing

TrigExpand[Tan[x + y]] /. Sin[u_] -> Cos[u] HoldForm[Tan[u]] //
   Simplify // ReleaseHold
(* (Tan[x] + Tan[y])/(1 - Tan[x] Tan[y]) *)

This works by making the (obviously valid) replacement

Sin[u_] -> Cos[u] Tan[u]

In itself, this would achieve little as (I guess) that Mathematica would simplify the result back to the unwanted form. However, wrapping HoldForm round the tangents means that Mathematica will not change them. Consequently, the simplest form is the one that we seek.

ReleaseHold removes the HoldForm wrapper. In this case, this doesn't change the appearance of the result, but means that the Tan terms will behave as expected in subsequent processing.