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:)
\[\tan 3a = \frac{ 3 \tan a - \tan³ a }{ 1 -3\tan² a }\]
I was thinking of starting with the LHS as tan (2a + a) but that runs into a dead end for me
No, focus on the more complicated side. :)
The RHS ressembles to this formula \[\tan 2a = \frac{ 2\tan a}{ 1- \tan² a}\] But as far as I know, if you multiply the a, you get tan x a, and not x tan a, so thats where i'm stuck
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solving by converting tan in terms of sin and cos will be much easier
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