Explanation

(c)
$\lim_{x \rightarrow \frac{\pi}{2}}\frac{\left( 1 - \tan\frac{x}{2} \right)(1 - \sin x)}{\left( 1 + \tan\frac{x}{2} \right)(\pi - 2x)^{3}}$
$= \lim_{h \rightarrow 0}\frac{1 - \tan\left( \frac{\pi}{4} - \frac{h}{2} \right)}{1 + \tan\left( \frac{\pi}{4} - \frac{h}{2} \right)}.\frac{(1 - \cos h)}{(2h)^{3}}$
$\left\lbrack let\ x = \frac{\pi}{2} - h\ as\ x \rightarrow \frac{\pi}{2},h \rightarrow 0 \right\rbrack$
$= \lim_{h \rightarrow 0}\tan\frac{h}{2}.\frac{2\sin^{2}\frac{h}{2}}{8h^{3}}\ \ \ \ \ \ \left\lbrack \because\tan{\left( \frac{\pi}{4} - x \right) = \frac{1 - \tan x}{1 + \tan x}} \right\rbrack$
$= \lim_{h \rightarrow 0}\frac{1}{4}.\frac{\tan\frac{h}{2}}{\frac{h}{2} \times 2} \times \left( \frac{\sin\frac{h}{2}}{\frac{h}{2}} \right)^{2} \times \frac{1}{4} = \frac{1}{32}$