Help with this wild limit.

$$\lim_{\alpha \rightarrow 5\pi/2}\frac{ \cos(\csc (2\alpha)+\cot(2\alpha))-1 }{ \tan(\frac{ \cos^2\alpha }{ 1+\sin \alpha }) }$$

So for more clarity ,it's the limit as alpha goes to 5pi/2 of the following quotien

numerator: $$\cos(\csc(2\alpha)+\cot(2\alpha))-1$$
denominator: $$\tan(\frac{ \cos^2\alpha }{ 1+\sin \alpha })$$

Question

Help with this wild limit.

$$\lim_{\alpha \rightarrow 5\pi/2}\frac{ \cos(\csc (2\alpha)+\cot(2\alpha))-1 }{ \tan(\frac{ \cos^2\alpha }{ 1+\sin \alpha }) }$$

So for more clarity ,it's the limit as alpha goes to 5pi/2 of the following quotien

numerator: $$\cos(\csc(2\alpha)+\cot(2\alpha))-1$$
denominator: $$\tan(\frac{ \cos^2\alpha }{ 1+\sin \alpha })$$

anonymous · Answer

yikes
i guess one first step would be to rewrite the denominator as 
$$\tan(1-\sin(\alpha))$$  maybe that might help

anonymous · Answer

bet we can clean up the argument inside the numerator as well 
i cheated and used the wolf, turns out tht 
$$\csc(2\alpha)+\cot(2\alpha)=\cot(\alpha)$$ although i don't see it right away, not familiar with "double angle" formulas for cosecant and cotangent

anonymous · Answer

at least rewriting is as 
$$\lim_{x\to \frac{5\pi}{4}}\frac{\cos(\cot(\alpha)-1}{\tan(1-\sin(\alpha))}$$ will make l'hopital less ugly
i wonder if there is some gimmick to make this more obvious