M Mahan1 Member Joined Oct 16, 2016 Messages 87 Gender Male HSC 2014 Feb 26, 2017 #2 well, the solution I came up with relies on induction, but I believe there is another way to solve it as well. Proof: prove the base case: n=2 Suppose it is true for n=k: we need to prove n=k+1: Adding to (*):
well, the solution I came up with relies on induction, but I believe there is another way to solve it as well. Proof: prove the base case: n=2 Suppose it is true for n=k: we need to prove n=k+1: Adding to (*):
 \ tan(2\alpha) = tan(\alpha)*\frac{2\ tan(\alpha)}{1-tan^{2}(\alpha)} = \frac{2tan^{2}(\alpha)}{1-tan^{2}(\alpha)} )
)+2}{1-tan^{2}(\alpha)} = -2 + \frac{tan(2 \alpha)}{tan(alpha)} = LHS )
\ tan(2\ \alpha) +....+tan((k-1)\ \alpha)\ tan(k\ \alpha) = \frac{tan(k \ \alpha)}{tan{\alpha}} - k \ \ (*))
\ tan(2\ \alpha) +....+tan(k\ \alpha)\ tan((k+1)\ \alpha) = \frac{tan((k+1) \ \alpha)}{tan{\alpha}} - (k+1) )
\ tan((k+1)\ \alpha) )
\ tan(2\ \alpha) +....+tan(k\ \alpha)\ tan((k+1)\ \alpha) = (\frac{tan(k \ \alpha)}{tan{\alpha}} - k) +tan(k\ \alpha)\ tan((k+1)\ \alpha))
}{tan{\alpha}} - k) +tan(k\ \alpha)\ tan((k+1)\ \alpha = \frac{tan(k \ \alpha)}{tan{\alpha}} +1 +tan(k\ \alpha)\ tan((k+1)\ \alpha -(k+1) )
\ tan((k+1)\ \alpha = \frac{tan((k+1)\ \alpha) - tan(k \ \alpha)}{tan(\alpha)} ))
}{tan{\alpha}} +\frac{tan((k+1)\ \alpha) - tan(k \ \alpha)}{tan(\alpha)} - (k+1) = \frac{tan((k+1)\ \alpha)}{tan(\alpha)} - (k+1) )
