2. The curve determined by the polar equation $\rho=\frac{1}{1-\cos \theta+\sin \theta}$ is
(A) Circle;
(B) Ellipse;
(C) Hyperbola;
(D) Parabola.
2. (C)
The original equation simplifies to $\rho=\frac{1}{1-\sqrt{2} \cos \left(\theta+\frac{\pi}{4}\right)}$.
Given the eccentricity $e=\sqrt{2}>1$, the curve determined is a hyperbola.
Another solution: The original equation can be transformed into $\rho-\rho \cos \theta+\rho \sin \theta=1$, which converts to the Cartesian equation
$$
\sqrt{x^{2}+y^{2}}=x-y+1,
$$

simplifying to
$$
2 x y-2 x+2 y=1 \text {. }
$$

That is, $(x+1)(y-1)=\frac{1}{2}$, indicating that the curve determined is a hyperbola.