$\displaystyle \lim_{x \to(\pi/2)^+}\frac{cosx}{1-sinx}$

I found that:

$\displaystyle \lim_{x \to(\pi/2)^+}cosx=0$

$\displaystyle \lim_{x \to(\pi/2)^+}1-sinx=0$

so I found the derivatives of them and I got:

$\displaystyle \lim_{x \to(\pi/2)^+}\frac{sinx}{cosx}=\frac{sin(\pi/2)}{cos(\pi/2)}=\frac{1}{0}$

I'm not sure on what I am doing wrong.