What is the eccentricity of the conic given by $r = \frac{2}{1 + \cos \theta - \sin \theta}$? A very simple question. Usually, Conic section in polar graph given as, 
$$r = \frac{ke}{1 - e \cos\theta}$$ 
or $\sin\theta$, or something along this way.  
However, given 
$$r = \frac{2}{1 + \cos \theta - \sin \theta}$$ 
how could I find its eccentricity?

You have 
\begin{align}
\cos\theta-\sin\theta&=-\sqrt2\,\left(-\frac1 {\sqrt2}\cos\theta+\frac1{\sqrt2}\sin\theta\right)\\ \ \\&=-\sqrt2\,\left(\cos\frac{3\pi}4\cos\theta+\sin\frac{3\pi}4\sin\theta\right)\\ \ \\ &
=-\sqrt2\,\cos\left(\theta-\frac{3\pi}4\right)
\end{align}
