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? 

HINT:
$$\frac{\sum_{n=a}^{b} \sin^2(n)}{\sum_{n=a}^{b} \cos^2(n)}=$$
$$\frac{2-2a+2b-\csc(1)\sin(1-2a)-\csc(1)\sin(1+2b)}{2-2a+2b+\csc(1)\sin(1-2a)+\csc(1)\sin(1+2b)}$$

Now we know that $a=1$ and $b=90$:
$$\frac{\sum_{n=1}^{90} \sin^2(n)}{\sum_{n=1}^{90} \cos^2(n)}=$$
$$\frac{2-2\cdot 1+2\cdot 90-\csc(1)\sin(1-2\cdot 1)-\csc(1)\sin(1+2\cdot 90)}{2-2\cdot 1+2\cdot 90+\csc(1)\sin(1-2\cdot 1)+\csc(1)\sin(1+2\cdot 90)}=$$
$$\frac{180-\csc(1)\sin(-1)-\csc(1)\sin(181)}{180+\csc(1)\sin(-1)+\csc(1)\sin(181)}=$$
$$\frac{182}{178}=\frac{91}{89}$$
