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)
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...