Find the following limit problem I'm trying to find the following limit 
$$ \lim_{x\to 0} \left({1-\sin x } {\cos x}\right)^{{ \csc 2x} }$$
How to prove the above limit equals $e^{-{1\over2}}$?

HINT:
$(1-\sin x \cos x)^{\csc 2x}=(1-\sin x \cos x)^{\frac{1}{\sin 2x}}=\left[(1-\sin x \cos x)^{\frac{1}{\sin x \cos }}\right]^{\sin x \cos x \cdot \frac{1}{\sin 2x}}=\left[(1-\sin x \cos x)^{\frac{1}{\sin x \cos }}\right]^{\sin x \cos x \cdot \frac{1}{2 \sin x \cos x}}=\left[(1-\sin x \cos x)^{\frac{1}{\sin x \cos }}\right]^{\frac{1}{2 }}$
