Question

Solve:
$\tan 3A=?$$tanA=\frac{4}{3}$

Answer 1

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

or $\tan 3A=\frac{3\times \frac{4}{3}-{\left(\frac{4}{3}\right)}^3}{1-3{\left(\frac{4}{3}\right)}^2}=\frac{\frac{108-64}{279}}{\frac{3-4}{3}}=\frac{44}{9}$

$\therefore \tan 3A=-\frac{44}{9}$ Ans