#### QUESTION

```
TrigExpand@Tan[x + y]
```

gives

$\frac{\sin (x) \cos (y)}{\cos (x) \cos (y)-\sin (x) \sin (y)}+\frac{\cos (x) \sin(y)}{\cos (x) \cos (y)-\sin (x) \sin (y)}$

but I expected

$\frac{\tan (x)+\tan (y)}{1-\tan (x) \tan (y)}$

Evaluating `TrigExpand`

with arguments $\tan (x-y)$ or $\tan (3*x)$ also returns results in terms of sine and cosine. What should I do to get results in terms of the tangent?

#### ANSWER

One way to induce *Mathematica* to simplify to `Tan`

functions is to introduce the arguments as inverse tangents, as in $x\equiv \arctan a$ and $y\equiv \arctan b$. Then you could write for example

```
Simplify[
TrigExpand@Tan[ArcTan[a] + ArcTan[b]]] /. {a -> Tan[x], b -> Tan[y]}
(* ==> (Tan[x] + Tan[y])/(1 - Tan[x] Tan[y]) *)
```

or more generally with an expression containing `x`

and `y`

:

```
expr = Tan[3 x];
Simplify[TrigExpand[
expr /. {x -> ArcTan[a], y -> ArcTan[b]}]] /. {a -> Tan[x],
b -> Tan[y]}
(* ==> (Tan[x] (-3 + Tan[x]^2))/(-1 + 3 Tan[x]^2) *)
```

**Edit**

Referring to the answer by @belisarius, I thought it's worth pointing out that you can achieve similar flexibility with my approach. For example, if you want the expression simplified in terms of `Sin`

instead of `Tan`

, just use inverse sines where I said to use inverse tangents above:

```
Simplify[TrigExpand[
expr /. {x -> ArcSin[a], y -> ArcSin[b]}]] /. {a -> Sin[x],
b -> Sin[y]}
(*
==> (Sin[x] (-3 + 4 Sin[x]^2))/(Sqrt[
1 - Sin[x]^2] (-1 + 4 Sin[x]^2))
*)
```

The insertion of the inverse function causes `Simplify`

to see the expression as more complex until it reaches a form in terms of the inverse variables. When the simplification is done, we revert the variables back without invoking `Simplify`

again.

You can try this with any target function that has an inverse.

Tweet