1. Given limit

We need to evaluate

$$L=\lim_{x\to \frac\pi2} \frac{\left(1-\tan\frac x2\right)\left(1-\sin x\right)}{\left(1+\tan\frac x2\right)(\pi-2x)^3}.$$

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2. Use the half-angle identity

Recall:

$$\tan\frac x2=\frac{\sin x}{1+\cos x}=\frac{1-\cos x}{\sin x}.$$

A very useful identity is

$$\frac{1-\tan\frac x2}{1+\tan\frac x2}=\tan\left(\frac\pi4-\frac x2\right).$$

Since

$$\frac\pi4-\frac x2=\frac{\pi-2x}{4},$$

we get

$$\frac{1-\tan\frac x2}{1+\tan\frac x2}=\tan\left(\frac{\pi-2x}{4}\right).$$

So the limit becomes

$$L=\lim_{x\to \frac\pi2} \tan\left(\frac{\pi-2x}{4}\right)\cdot \frac{1-\sin x}{(\pi-2x)^3}.$$

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3. Substitute a small variable

Let

$$h=\pi-2x.$$

Then as $x\to \frac\pi2$, we have $h\to 0$. Also,

$$x=\frac{\pi-h}{2}=\frac\pi2-\frac h2.$$

Hence

$$\sin x=\sin\left(\frac\pi2-\frac h2\right)=\cos\frac h2.$$

Therefore,

$$1-\sin x=1-\cos\frac h2.$$

So

$$L=\lim_{h\to 0} \tan\frac h4\cdot \frac{1-\cos\frac h2}{h^3}.$$

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4. Use standard small-angle expansions

As $h\to 0$,

$$\tan\frac h4 \sim \frac h4,$$

and

$$1-\cos\frac h2 \sim \frac{1}{2}\left(\frac h2\right)^2=\frac{h^2}{8}.$$

Thus,

$$\tan\frac h4\cdot (1-\cos\tfrac h2) \sim \frac h4\cdot \frac{h^2}{8}=\frac{h^3}{32}.$$

Therefore,

$$L=\lim_{h\to 0} \frac{h^3/32}{h^3}=\frac{1}{32}.$$

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5. Check options

The correct value is

$$\boxed{\frac{1}{32}}.$$

So the correct option is D.

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6. Comparison with stored answer

Stored correct answer: D

Our derived answer: D

So they agree.