Answer
$0$
Work Step by Step
$\lim _{x\rightarrow \dfrac {\pi }{2}}\dfrac {\dfrac {1}{\sqrt {\sin x}}-1}{x+\dfrac {\pi }{2}}=\dfrac {\dfrac {1}{\sqrt {\sin \dfrac {\pi }{2}}}-1}{\dfrac {\pi }{2}+\dfrac {\pi }{2}}=\dfrac {\dfrac {1}{\sqrt {1}}-1}{\pi }=\dfrac {1-1}{\pi }=0$
