Trigonometric identity, possible error 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?

Since all the trig values are squared, it seems as though the exercise is simply playing with shifts by odd or even multiples of $\pi/2$.
Loosely,

  
*
  
*Shifting by "$\frac{\pi}{2} \cdot \text{odd}$" switches "sin" and "cos" (and possibly affects the sign)
  
*Shifting by "$\frac{\pi}{2} \cdot \text{even}$" ($=$ "$\pi \cdot \text{any}$") preserves "sin" and "cos" (and possibly affects sign)
  
*Negating the argument preserves "sin" and "cos" (and possibly affects sign)
  

Since squaring eliminates sign considerations, we have, simply:

$$\begin{align}
\mathrm{trig}^2\left( \pm \; \theta \pm \frac{\pi}{2} \text{odd} \right) &= \mathrm{cotrig}^2\theta \\
\mathrm{trig}^2\left( \pm \; \theta \pm \frac{\pi}{2} \text{even} \right) &= \mathrm{trig}^2\left( \pm \; \theta \pm \pi \cdot \text{any} \right) = \mathrm{trig}^2\theta
\end{align}$$

where each "$\pm$" is independent, "any" means (of course) "any integer", and "trig" can in fact be any of the six trig functions.
This makes pretty quick work of the simplification process ...
$$\begin{align}
\frac{\sin^2\left(\frac{5\pi}{6}-\alpha\right)}{\cos^2\left(\alpha-4\pi\right)} - \cot^2\left(\alpha-11\pi\right) \; \sin^2\left(-\alpha-\frac{13\pi}{2}\right) &\stackrel{?}{=} \sin^2\alpha \\[1em]
\frac{\sin^2\left(\frac{5\pi}{6}-\alpha\right)}{\cos^2\alpha} - \cot^2\alpha \; \cos^2\alpha &\stackrel{?}{=} \sin^2\alpha
\end{align}$$
... right up to the point at which the process shudders to a halt.
Given the nature of all the other terms (and @Adam's comment that sum and difference identities are not allowed), I suspect that "$\frac{5\pi}{6}$" is a typo of "$\frac{5\pi}{2}$", which would get us a little further ...
$$\frac{\cos^2\alpha}{\cos^2\alpha} - \cot^2\alpha \; \cos^2\alpha = 1 - \cot^2\alpha \;\cos^2\alpha \stackrel{?}{=} \sin^2\alpha$$
... but we hit another snag. Could it be that "$\sin^2\left(-\alpha-\frac{13\pi}{2}\right)$" is a typo of "$\cos^2(...)$"? If so, then that factor should've simplified to "$\sin^2\alpha$", and we'd have
$$1 - \cot^2\alpha \;\sin^2\alpha = 1 - \cos^2\alpha = \sin^2\alpha$$
as desired.
(It's also possible that, instead of a sin-cos typo, "$\cot$" is a typo for "$\tan$", but it seems like that would be an easier one for the OP to notice.)
