Answer
$\dfrac{1}{4}$
Work Step by Step
Given: $\lim\limits_{\theta\to\pi/2}\frac{1-\sin\theta}{1+\cos2\theta}$
Here, $\lim_{\theta\to\pi/2}(1-\sin\theta)=1-\sin(\frac{\pi}{2})=1-1=0$ and $\lim_{\theta\to\pi/2}(1+\cos2\theta)=1+\cos(2\times\frac{\pi}{2})=1+\cos\pi=1-1=0,$
This shows an indeterminate form of type $\dfrac{0}{0}$, and so we will apply L'Hospital Rule.
$\lim\limits_{\theta\to\pi/2}\dfrac{\frac{d}{d\theta}(1-\sin\theta)}{\dfrac{d}{d\theta}(1+\cos2\theta)}=\lim\limits_{\theta\to\pi/2}\dfrac{-\cos\theta}{-2\sin2\theta}=\lim\limits_{\theta\to\pi/2}\frac{\cos\theta}{2\sin2\theta}$
Now, again with this limit this shows an indeterminate form of type $\dfrac{0}{0}$, and so we will apply L'Hospital Rule.
Thus,
$=\lim\limits_{\theta\to\pi/2}\dfrac{-\sin\theta}{4\cos2\theta}$
$=\dfrac{-\sin(\pi/2)}{4\cos(2\times\pi/2)}$
or, $=\dfrac{-1}{4\cos\pi}$
or, $=\dfrac{-1}{4(-1)}$
or, $ =\dfrac{1}{4}$
