Help mi plzz.​

Tan3A=tan(2A+A)

We know that , tan(x+y)=(tanx + tany)/(1 - (tanx)(tany))

tan(2A+A)=(tan2A+tanA)/(1 - (tan2A)(tanA))— (1)

We know that , tan2x=2tanx/(1 - tan^2x)

So, by substituting tan2A in (1),we get,

=[2tanA/(1 - tan^2A) + tanA]/1- (2tanA/(1 - tan^2A))(tanA)]

=[2tanA+tanA - tan^3A]/[1 - tan^2A - 2tan^2A]

=[3tanA - tan^3A]/[1-3tan^2A]

Therefore, tan3A= [3tanA - tan^3A]/[1-3tan^2A]