4. Let $a=\frac{1}{2} \cos 6^{\circ}-\frac{\sqrt{3}}{2} \sin 6^{\circ}$,
$$
b=\frac{2 \tan 13^{\circ}}{1-\tan ^{2} 13^{\circ}}, c=\sqrt{\frac{1-\cos 50^{\circ}}{2}} \text {. }
$$

Then we have ().
(A) $ab>c$
(D) $b<c<a$
4.A.

Since $a=\frac{1}{2} \cos 6^{\circ}-\frac{\sqrt{3}}{2} \sin 6^{\circ}=\sin 24^{\circ}$,
$$
\begin{aligned}
b & =\frac{2 \tan 13^{\circ}}{1-\tan ^{2} 13^{\circ}}=\tan 26^{\circ}, \\
c & =\sqrt{\frac{1-\cos 50^{\circ}}{2}}=\sin 25^{\circ},
\end{aligned}
$$

Therefore, $a<c$.
Also, for $x \in\left(0, \frac{\pi}{2}\right)$, we have $\sin x<x<\tan x$. Hence $\sin 25^{\circ}<\sin 26^{\circ}<\tan 26^{\circ}$, which means $c<b$.