Prove that, $\tan 3 \mathrm{~A}=\frac{3 \tan \mathrm{A}-\tan ^{3} \mathrm{~A}}{1-3 \tan ^{2} \mathrm{~A}}$.

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written 3.0 years ago by

Krishna_Agrawal • 2.7k

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,

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

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

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

$Hence, Proved.$

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