Question

Prove:
$\tan ({60}^o-A)\tan A\tan ({60}^o+A)=\tan 3A$

Answer 1

$\tan ({60}^o-A)\tan A\tan ({60}^o+A)=\tan 3A$

LHS $=\frac{\tan {60}^o-\tan A}{1+\tan {60}^o\tan A}\times \tan A\times \frac{\tan {60}^o+\tan A}{1-\tan {60}^o\tan A}$

$=\frac{{\tan }^2{60}^o-{tan}^{2}A}{1-{\tan }^2{60}^o{tan}^{2}A}\times \tan A$

$=\frac{{\left(\sqrt{3}\right)}^2-{\tan }^{2}A}{1-{\left(\sqrt{3}\right)}^2{\tan }^{2}A}\times \tan A$

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