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thank you so much guys!

yea my bad, tanθtan2θ+tan2θtan3θ+...tannθtan(n+1)θ=tan(n+1)θcotθ-(n+1) I got the base case, but after idek how to sub in the inductive step its worth mentioning there is a part a which gives the identity tana + tanb = tan(a+b) * (1-tanatanb)

Gonna reopen this thread after a while lol, so how do you prove the second and third steps?

@Luukas.2 do you have any insights?

I was trying to do this question too, all I was able to do for c was find the sums of couples of roots, (ab + cd + ae) and make them equal -3/2, then one of those terms is -cos π /12 * cos 5π/12 and using my calculator to compute the other terms i got that value as 1/4 but i feel there should be...

a) fully factorise z^6-1=0 over the rational field b) solve z^2+z^4+z^6 over the complex field how do you use part a to do part b?