# Tan Summation 2

Geometry Level 4

Let $$\displaystyle a= \sum _{r=1}^{11} \tan^2 \left ( \dfrac{r \pi}{24} \right )$$ and $$\displaystyle b= \sum _{r=1}^{11} (-1)^{r-1}\tan^2 \left ( \dfrac{r \pi}{24} \right )$$.

Then find the value of $$\dfrac{(2a-b)-(2b-a)}{log_{2b-a} {(2a-b)}}$$.

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