Can $\frac{\csc \alpha +\cos \alpha}{\cos \alpha - \tan \alpha - \sec \alpha}$ be simplified? I am trying to simplify the following but I cannot.
$$
\frac{\csc \alpha +\cos \alpha}{\cos \alpha - \tan \alpha - \sec \alpha}
$$
Can it be simplified?
Edit
My last result is
$$
- \frac{\cos \alpha \left( 1 + \sin \alpha \cos \alpha\right)}
{\sin^2 \alpha \left(1+\sin \alpha\right)}
$$
I am wondering it might be a wrong question given by my student's teacher.
If you multiply the numerator and denominator of your last expression by $4(1-\sin\alpha)$ and use the identity $\sin2\alpha=2\sin\alpha\cos\alpha$, you can rewrite your result as $$(1-\csc\alpha)(2\csc2\alpha+1).$$ This reduces the original 24 symbols to 17.
First, let us state of the fundamentals.
$$\csc \alpha = \frac{1}{\sin \alpha};\ \tan \alpha = \frac{\sin \alpha}{\cos \alpha};\ \cot \alpha = \frac{\cos\alpha}{\sin \alpha};\ \sec \alpha = \frac{1}{\cos \alpha};\ \sin^2\alpha + \cos^2\alpha = 1;$$

So,
$$\frac{\csc \alpha +\cos \alpha}{\cos \alpha - \tan \alpha - \sec \alpha} \cdot 1= \frac{\frac{1}{\sin \alpha} + \cos \alpha}{\cos \alpha - \frac{\sin\alpha}{\cos\alpha} - \frac{1}{\cos\alpha}} \cdot\frac{\cos \alpha}{\cos \alpha} = \frac{\cot \alpha + \cos^2\alpha}{\cos^2\alpha - \sin\alpha - 1} = \frac{\cot\alpha + 1 -\sin^2\alpha}{1-\sin^2\alpha - \sin\alpha - 1} = \frac{\cot\alpha + 1 -\sin^2\alpha}{\left(0-\sin\alpha\right)\cdot\left(1+\sin\alpha\right)}\cdot\frac{\sin\alpha}{\sin\alpha} = \frac{\cos\alpha + \sin\alpha - \sin^3\alpha}{- \sin^2\alpha\left(1+\sin\alpha\right)} = -\frac{\cos\alpha + \sin\alpha\cos^2\alpha}{\sin^2\alpha\left(1+\sin\alpha\right)} = - \frac{\cos\alpha\left(1 + \sin\alpha\cos\alpha\right)}{\sin^2\alpha\left(1+\sin\alpha\right)}$$
Alright, a great number of further simplifications are possible from here, but what do you think will eliminate the denominator and bring the order down to one without changing the inner $\alpha$. There are 4 terms and various manipulations left.

You'll see this is actually the simplest you can get it to. It is not trivial to prove this though.