What's the Trick to Proving tan3A=3tanA-tan^3 A/ 1-3tan^2 A?

[rhule009]

can anyone help[ me prove the following identity i keep on ending up in a dead end

tan3A=3tanA-tan^3 A/ 1-3tan^2 A
thank you

 

[courtrigrad]

you know that \tan(A+B) = \frac{\tan A + \tan B}{1-\tan A \tan B}. Setting A = B we get \tan 2A = \frac{2\tan A}{1-\tan A \ tan A}. So what is \tan(A+2A)? Thus work from the left to the right.

 

[rhule009]

i end up with
tanA+2tanA/1-tanAtanAtanA

 

[rhule009]

what am i doing wrong