\(\lim _{x \rightarrow \frac{\pi}{2}} \frac{\left(1-\tan \left(\frac{x}{2}\right)\right)(1-\sin x)}{\left(1+\tan \left(\frac{x}{2}\right)\right)(\pi-2 x)^3}\) is

\(\begin{aligned}
\text {Let } l & =\lim _{x \rightarrow \frac{\pi}{2}}\left[\frac{1-\tan \left(\frac{x}{2}\right)}{1+\tan \left(\frac{x}{2}\right)}\right]\left[\frac{(1-\sin x)}{(\pi-2 x)^3}\right] \\
& =\lim _{x \rightarrow \frac{\pi}{2}} \frac{\left.\tan \left(\frac{\pi}{4}-\frac{x}{2}\right)\right](1-\sin x)}{(\pi-2 x)^3}
\end{aligned}\)
Put \(\pi-2 x=\theta\)
\(\Rightarrow x=\frac{\pi}{2}-\frac{\theta}{2}\) and as \(x \rightarrow \frac{\pi}{2}, \theta \rightarrow 0\)
\(\begin{aligned} \therefore \quad l & =\lim _{\theta \rightarrow 0} \frac{\tan \frac{\theta}{4}\left(1-\cos \frac{\theta}{2}\right)}{\theta^3} \\ & =\lim _{\theta \rightarrow 0} \frac{\tan \frac{\theta}{4}}{\frac{\theta}{4} \times 4} \cdot \frac{2 \sin ^2 \frac{\theta}{4}}{\frac{\theta^2}{16} \times 16} \\ & =\frac{1}{32} \lim _{\theta \rightarrow 0}\left[\frac{\tan \frac{\theta}{4}}{\frac{\theta}{4}} \cdot\left(\frac{\sin \frac{\theta}{4}}{\frac{\theta}{4}}\right)^2\right]=\frac{1}{32}(1)(1)^2=\frac{1}{32}\end{aligned}\)