#### Answer

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

#### Work Step by Step

$$A=\lim_{x\to(\pi/2)^+}\frac{\cos x}{1-\sin x}$$
In this exercise, the elementary method seems to lead nowhere. Therefore, we will only try applying L'Hospital's Rule.
Since $\lim_{x\to(\pi/2)^+}(\cos x)=\cos(\frac{\pi}{2})=0$ and $\lim_{x\to(\pi/2)^+}(1-\sin x)=1-\sin(\frac{\pi}{2})=1-1=0$,
this limit is an indeterminate form of $\frac{0}{0}$, so we can apply L'Hospital's Rule.
$$A=\lim_{x\to(\pi/2)^+}\frac{\frac{d}{dx}(\cos x)}{\frac{d}{dx}(1-\sin x)}$$
$$A=\lim_{x\to(\pi/2)^+}\frac{(-\sin x)}{0-\cos x}$$
$$A=\lim_{x\to(\pi/2)^+}\frac{\sin x}{\cos x}$$
As $x$ approaches $\frac{\pi}{2}$ from the right side, $\sin x$ approaches $1$, but $\cos x$ approaches $0$. A glance look tells us the result of this limit should be $\infty$.
However, as $x$ approaches $\frac{\pi}{2}$ from the right side, which means $x$ is close to the point $\frac{\pi}{2}$, but still $x\gt\frac{\pi}{2}$, we have $$\sin x\gt0\hspace{1cm}and\hspace{1cm}\cos x\lt0$$
Therefore, this limit is $-\infty$ for the numerator is positive but the denominator is negative.