I need to prove the following trigonometric identity: $$ \frac{\sin^2(\frac{5\pi}{6} - \alpha )}{\cos^2(\alpha - 4\pi)} - \cot^2(\alpha - 11\pi)\sin^2(-\alpha - \frac{13\pi}{2}) =\sin^2(\alpha)$$
I cannot express $\sin(\frac{5\pi}{6}-\alpha)$ as a function of $\alpha$. Could it be a textbook error?