<s>[INST] Sine/cosine series $$\frac{\sin²(1°) + \sin²(2°) + \sin²(3°) + .. + \sin²(90°)}{\cos²(1°) + \cos²(2°) + \cos²(3°) + .. + \cos²(90°)} = ?$$
I tried to use multiple identities but I couldn't simplify the expression. Where should I start? 
 [/INST] $$\frac{( sin²(1) + sin²(2) + sin²(3) + .. + sin²(90) )}{ ( cos²(1) + cos²(2) +cos²(3) + .. + cos²(90) )} = \\ \frac{2}{2}*\frac{( sin²(1) + sin²(2) + sin²(3) + .. + sin²(90) )}{ ( cos²(1) + cos²(2) +cos²(3) + .. + cos²(90°) )}$$ so try by $sin^2x=\frac{1-cos2x}{2} \\cos^2x=\frac{1-cos2x}{2}$
so 
$$\frac{1-cos2 +1-cos 4+1-cos 6+ ...+1-cos 180}{1+cos2 +1+cos 4+1+cos 6+ ...+1+cos 180}=\\ \frac{90-(cos 2+cos 4+cos 6+ ...+cos 180)}{90+(cos 2+cos 4+cos 6+ ...+cos 180)}$$ 
now see that $$(cos 2+cos 4+cos 6+ ...+cos 180)=\\(cos 2+cos 178 )+(cos 4 +cos 176) +...(cos 44 +cos 46) +cos 90 +cos 180 =0 +0 +0+0+...+cos 90 +cos 180 =0+0+(-1)$$ so  $$\frac{90-(cos 2+cos 4+cos 6+ ...+cos 180)}{90+(cos 2+cos 4+cos 6+ ...+cos 180)}=\frac{90-(-1)}{90+(-1)}=\frac{91}{89}$$
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