Express tan3θ in terms of tanθ

That's how I would do it. Is it off of STEP I 05?

Is that the right way to do this sort of question, or is there a quicker way to do it?

I managed to get the right answer for this question: {Tanθ(3-tan²&#952}/(1-3tan²&#952 By writing it as tan(2θ + &#952, expanding it using the compound formula, then expanding the tan2θ terms using the double angle formula and then simplifying (which took quite a few steps).

Is that the right way to do this sort of question, or is there a quicker way to do it?

I think this is quicker:

Start with the identities for sin3x/cos3x

=(3sinx-4sin³x)/(4cos³x-3cosx)

divide throughout by cos³x to get

(3tanx.sec²x-4tan³x)/(4-3sec²x)

replace the 2 instances of sec²x with (1+tan²x)

get (3tanx -tan^3x)/(1-3tan^2x) as required.

Sorry about all the ^ use! Perhaps we'll get the toolbar back soon...

Hmm, thanks. That way looks quicker but knowing me I'd probably make mistakes substituting things in, lol.

Btw you can use {sup}blah{/sup} (with square brackets) to make text superscript.