روش دوم: $\lim_{x \to (\pi/2)^+} \frac{\cos x}{1 - \sin x} \times \frac{1 + \sin x}{1 + \sin x} = \lim_{x \to (\pi/2)^+} \frac{\cos x(1 + \sin x)}{1 - \sin^2 x} = \lim_{x \to (\pi/2)^+} \frac{\cos x(1 + \sin x)}{\cos^2 x} = \lim_{x \to (\pi/2)^+} \frac{1 + \sin x}{\cos x} = \frac{2}{0^-} = -\infty$