$\frac{1+\cos 56°+\cos 58°-\cos 66°}{\cos 28°\cos 29°\sin 33°}=$

Let $A=56,B=58,C=66\text{and}A+B+C=180°$

$\Rightarrow \cos A+\cos B-\cos C=2\cos \frac{A+B}{2}.\cos \frac{A-B}{2}-1+2{\sin }^2\frac{C}{2}$

$=2\sin \frac{C}{2}\left(\cos \frac{A-B}{2}+\cos \frac{A+B}{2}\right)-1$

$=2\sin \frac{C}{2}\left(2\cos \frac{A}{2}.\cos \frac{B}{2}\right)-1$

$=4\cos \frac{A}{2}.\cos \frac{B}{2}.\sin \frac{C}{2}-1$

Consider $\frac{1+\cos 56°+\cos 58°-\cos 66°}{\cos 28°.\cos 29°.\sin 33°}$

$=\frac{1+4\cos \frac{56}{2}.\cos \frac{58}{2}.\sin \frac{66}{2}-1}{\cos 28°.\cos 29°.\sin 33°}$

$=\frac{4\cos 28°.\cos 29°.\sin 33°}{\cos 28°.\cos 29°.\sin 33°}=4$