Let S={x ∈(-(π)/2, (π)/2): 9^1- tan^2 x+9^tan^2 x=10} and β=Σ_x ∈ s tan^2(x/3), then 1/6(β-14)^2 is equal to

Let $\displaystyle S=\left\{x \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right): 9^{1-\tan ^2 x}+9^{\tan ^2 x}=10\right\}$ and $\displaystyle \beta=\sum_{x \in s} \tan ^2\left(\frac{x}{3}\right)$, then $\displaystyle \frac{1}{6}(\beta-14)^2$ is equal to