Tan3x=3Tanx-Tan^3x by 1-3tan^2x Prove it?

We have, `tan(x+y)=(tanx+tany)/(1-tanxtany)............(diamond)`.

Letting `x=y=A`, we get,

`tan(A+A)=(tanA+tanA)/(1-tanA*tanA)`.

`:. tan2A=(2tanA)/(1-tan^2A)............(diamond_1)`.

Now, we take, in `(diamond), x=2A, and, y=A`.

`:. tan(2A+A)=(tan2A+tanA)/(1-tan2A*tanA)`.

`:. tan3A={(2tanA)/(1-tan^2A)+tanA}/{1-(2tanA)/(1-tan^2A)*tanA}`,

`={(2tanA+tanA(1-tan^2A))/(1-tan^2A)}-:{1-(2tan^2A)/(1-tan^2A)}`,

`=(2tanA+tanA-tan^3A)/(1-tan^2A-2tan^2A)`.

`rArr tan3A=(3tanA-tan^3A)/(1-3tan^2A)`, as desired!

Let's do it from first principles from De Moivre:

`cos 3 x + i sin 3x = (cos x + i sin x)^3`

Using the `1,3,3,1` row of Pascal's triangle,

`cos 3 x + i sin 3x`

`= cos^3 x + 3 \cos^2 x (i \sin x) + 3 \cos x (i^2 \sin^2 x ) + i^3 sin^3 x`

`= (cos^3 x- 3 \cos x \sin^2 x ) + i ( 3 \cos^2 x \sin x - sin^3 x)`

Equating respective real and imaginary parts,

`\cos 3 x = cos^3 x- 3 \cos x \sin^2 x`

`\sin 3x = 3 \cos ^2 x \sin x - \sin ^3 x`

Those are (a fairly obscure form of) the triple angle formulas, and typically we'd just write those or a more standard form down and start from here.

`\tan 3x = \frac{ sin 3x}{cos 3x } = frac{ 3 \cos ^2 x \sin x - \sin ^3 x}{ cos^3 x- 3 \cos x \sin^2 x} cdot \frac{1/cos^3 x}{1/ cos^3 x}`

`tan 3x = \frac{3 tan x - tan^3 x}{1 - 3 tan ^2 x} \quad \square`