Simplifying $\cot(\beta - \gamma)\cot(\gamma - \alpha) + \cos(\gamma - \alpha)\cos(\alpha - \beta) + \cos(\alpha - \beta)\cot(\beta - \gamma) $ 
I have a homework question which asks to simplify the following expression:
$$\cot(\beta - \gamma)\cot(\gamma - \alpha) + \cos(\gamma - \alpha)\cos(\alpha - \beta) + \cos(\alpha - \beta)\cot(\beta - \gamma) $$

There are no restrictions placed on $\alpha, \beta, \gamma$ except that they take real values.
I tried to solve the question by writing $\cot$ as $\dfrac{\cos}{\sin}$ and using product to sum formulae but instead of being simplified the expression only became more humongous.
How should I proceed with the question? Any help or hint will be appreciated.
Edit:
The expression as stated is incorrect and has no further simplification, as told by Somos. I confirmed it with my teacher. The cosines should be cotangents, which is simplified as told by Lab Bhattacharjee.
 A: It is a priori difficult to solve "simplify" problems because there is no precise standard definition of how one expression is simpler than another. Even if there were such a definition, it is difficult to prove that an expression is already as simple as possible without some theory. Despite this, many computer algebra systems implement some form of  a "simplify" command.
For your given expression
$$ \cot(\beta - \gamma)\cot(\gamma - \alpha) + \cos(\gamma - \alpha)\cos(\alpha - \beta) + \cos(\alpha - \beta)\cot(\beta - \gamma) \tag1 $$
it seems that this can not be simplified according to Mathematica. As a sum of products of three pairs of trigonometric functions it is about as simple as it can be. Given this, it is possible that there was a misprint and some of the $\cos$ should be $\cot$ or vice versa.
If the last $\cos$ was changed to $\cot$, then there is a factorization
$$ \cot(\beta - \gamma)\cot(\gamma - \alpha) + \cos(\gamma - \alpha)\cos(\alpha - \beta) + \cot(\alpha - \beta)\cot(\beta - \gamma) = (\cos(\beta-\gamma) + \sin(2(\alpha-\beta))/4) \frac{ \sin(2(\alpha-\gamma)) }
{\sin(\alpha-\beta) \sin(\alpha-\gamma) } \tag2 $$
but it does not seem that the product is simpler than the triple sum.
If all of the $\cos$ were changed to $\cot$ then the result
$$ \cot(\beta - \gamma)\cot(\gamma - \alpha) + \cot(\gamma - \alpha)\cot(\alpha - \beta) + \cot(\alpha - \beta)\cot(\beta - \gamma) = 1 \tag3 $$
is as simple as you can get except for $0$. The question has severe problems and is almost certain that it was incorrectly stated.
