Let $$a=\sum _{r=1}^{11}tan^2\bigg(\frac{r\pi}{24}\bigg) $$and $$b=\sum _{r=1}^{11}(-1)^{r-1}tan^2\bigg(\frac{r\pi}{24}\bigg) $$ then find the value of $log_{(2b-a)}{(2a-b)}$.
In $a$ I find that $$tan^2\bigg(\frac{\pi}{24}\bigg)+tan^2\bigg(\frac{11\pi}{24}\bigg)=tan^2\bigg(\frac{\pi}{24}\bigg)+cot^2\bigg(\frac{\pi}{24}\bigg) =\frac{2}{sin^2\big(\frac{\pi}{12}\big)}-2$$ How do I use this to solve the problem ? How do I proceed further ?
