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?
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.
On
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.
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.