Evaluating $ \lim_{x \to 0}{ \arctan( 2 {\Large[} \frac{\cos(x) - 1}{\sin^2x} {\Large]} )} $ Having 
$$ \lim_{x \to 0}{ \arctan\left( 2 \left[ \frac{\cos(x) - 1}{\sin^2x}\right] \right)} = L $$
The interesting part however is $$ \frac{\cos(x) - 1}{\sin^2x} $$ and $ \lim_{x \to 0} \frac{(\cos(x) - 1)}{\sin^2x} = \left[\frac{0}{0}\right] = -1$. With l'hopital's rule it's easy to solve - I'm interested in other methods though1. Hints are welcome too.

Attempt 1 
Considering $$ 1 = \sin^2x + \cos^2x $$ Plugging in:
 $$ \frac{\cos(x) - \sin^2x - \cos^2x}{\sin^2x} 
\longrightarrow \lim_{x \to 0} {\cot(x)\csc(x) - 1 - \cot^2x}$$
which doesn't go very far.
Attempt 2
Considering:
$$ -2 \leq \cos(x) - 1 \leq 0 \rightarrow \frac{-2}{\sin^2x} \leq \cos(x) - 1 \leq \frac{0}{\sin^2x}$$ 
But $$ \lim_{x\to0}{  \frac{-2}{\sin^2x} } \neq \lim_{x\to0}{ \frac{0}{\sin^2x}} $$
so the squeeze theorem cannot be applied. (bonus question: can it?)
Please avoid Taylor expansion.
 A: $${\cos x-1\over\sin^2x}={\cos x-1\over\sin^2x}\cdot{\cos x+1\over\cos x+1}={\cos^2x-1\over\sin^2x(\cos x+1)}={-\sin^2x\over\sin^2x(\cos x+1)}={-1\over\cos x+1}$$
A: Hint :
$$\lim_{x\to 0}\frac{\sin^2x}{x^2}=1$$
$$\lim_{x\to 0}\frac{1-\cos x }{x^2}=\frac12$$
A: $$\frac{\cos x-1}{\sin^2 x}=\frac{\cos x-1}{x^2}\frac{x^2}{\sin^2 x}$$
Both $\frac{1-\cos x}{x^2}$ and $\frac{\sin x}{x}$ are basic limits.
A: Note that $\cos(x)-1=-2\sin^2(x/2)$.  Therefore, we have
$$\begin{align}
\frac{\cos(x)-1}{\sin^2(x)}&=-\frac{2\sin^2(x/2)}{\sin^2(x)}\\\\
&=-\frac12 \left(\left(\frac{\sin(x/2)}{(x/2)}\right)\,\left(\frac{x}{\sin(x)}\right)\right)^2
\end{align}$$
A: Use:


*

*$$\frac{\cos(x)-1}{\sin^2(x)}=-\frac{\sec^2\left(\frac{x}{2}\right)}{2}$$

*$$\arctan(-x)=-\arctan(x)$$


So, we get:
$$\arctan\left(2\cdot\frac{\cos(x)-1}{\sin^2(x)}\right)=\arctan\left(-\sec^2\left(\frac{x}{2}\right)\right)=-\arctan\left(\sec^2\left(\frac{x}{2}\right)\right)$$
A: Hint:
$$\dfrac{\cos x-1}{\sin^2x}=-\dfrac1{1+\cos x}$$ for $\cos x-1\ne0$
and
$$\arctan(-x)=-\arctan x$$
A: $(\cos x - 1) / \sin ^2x $
$= (\cos x- 1) / (1 - \cos^2x)$
$ = (\cos x - 1) / [(1 - \cos )(1 + \cos )]$
$ = - 1 / (\cos x + 1)$
