$\because \text{R.H.L.}=\underset{h\to 0^{+}}{lim}f(\frac{\pi }{2}+h)=\underset{h\to 0^{+}}{lim}\frac{\sin (-\sinh )}{h}\to \mathrm{\infty }$

$\because \text{L.H.L.}=\underset{h\to 0^{-}}{lim}f(\frac{\pi }{2}-h)=\underset{h\to 0}{lim}\frac{\sin \left(\sinh \right)}{-h}$

$=\underset{h\to 0^{+}}{lim}(\frac{\sin \left(\sinh \right)}{\sinh }\times \frac{\left(\sinh \right)}{-h})=1\times -1=-1$

$\therefore \text{L.H.L}\ne \text{R.H.L}$