To Proof :-

$$\tan 3 \mathrm{~A}=\frac{3 \tan \mathrm{A}-\tan ^{3} \mathrm{~A}}{1-3 \tan ^{2} \mathrm{~A}}$$

Proof :-

As we know -

$$\tan(A+B) = \frac{tan\ A\ +\ tan\ B}{1\ - tan\ A\ tan\ B}$$

Put B = 2A.

$$\tan A\ +\ 2A\ = \frac{tan\ A\ +\ tan\ 2A}{1\ - tan\ A\ tan\ 2A}$$

As we know,

$$\tan\ 2A\ =\ \frac{2\ tan\ A}{1\ -\ tan^2\ A}$$

So,

$$\tan\ 3A\ = \frac{tan\ A + \frac{2\ tan\ A}{1\ -\ tan^2\ A}}{1\ -\ tan\ A \frac{2\ tan\ A}{1\ -\ tan^2\ A}}$$

$$\tan\ 3A\ =\ \frac{\frac{tan\ A - tan^3\ A + 2tan\ A}{1\ -\ tan^2\ A}}{\frac{1\ -\ tan^2\ A\ - 2tan^2\ A}{1\ -\ tan^2\ A}}$$

$$\tan\ 3A\ =\ \frac{3 \tan \mathrm{A}-\tan ^{3} \mathrm{~A}}{1-3 \tan ^{2} \mathrm{~A}}$$

$Hence, Proved. $