Short method to evaluate $\lim_{x\to \frac\pi2} \frac{(1-\tan(\frac x2))(1-\sin(x))}{(1+\tan(\frac x2))(\pi-2x)^3}$? $$\lim_{x\to \frac\pi2} \frac{(1-\tan(\frac x2))(1-\sin(x))}{(1+\tan(\frac x2))(\pi-2x)^3}$$
I only know of L'hopital method but that is very long. Is there a shorter method to solve this?

Another trick is to multiply both the numerator and denominator by 
$(1+\tan(x/2))(1+\sin x)$ and use that
\begin{align}
1-\sin^2x&=\cos^2x,\\
1-\tan^2(x/2)&=\frac{\cos^2(x/2)-\sin^2(x/2)}{\cos^2(x/2)}=\frac{\cos x}{\cos^2(x/2)}.
\end{align}
Then you get after simplification
$$
\frac{1}{(1+\tan(x/2))^2(1+\sin x)\cos^2(x/2)}\frac{1}{2^3}\color{red}{\left(\frac{\cos x}{\pi/2-x}\right)^3}
$$
where the only uncertainty  when $x\to\pi/2$ is in the red part. The limit of $\frac{\cos x}{\pi/2-x}$ can be calculated by L'Hopital.
