Solution:

Given, \(\lim_{x\rightarrow\frac{\mathrm\pi}2}\frac{\cos x}{{\displaystyle\frac{\mathrm\pi}2}-x}\)

= \(\lim_{x\rightarrow\frac{\mathrm\pi}2\;\;\;\;\;\;}\left(\frac{\sin\left({\displaystyle\frac{\mathrm\pi}2}-x\right)}{\left(\frac{\mathrm\pi}2-x\right)}\right)\)

= \(\lim_{x\rightarrow\frac{\mathrm\pi}2\;\;\;\;\;\;}\left(\frac{-\sin\left(x-{\displaystyle\frac{\mathrm\pi}2}\right)}{-\left(x-\frac{\mathrm\pi}2\right)}\right)\)

= 1