Calculate without using tables: $\;\;\;$ $\dfrac{\sin 110^\circ \sin 250^\circ + \cos 540^\circ \cos 290^\circ \cos 430^\circ}{\cos^2 1260^\circ}$
$\sin 110^\circ = \sin \left(90^\circ + 20^\circ\right) = \cos 20^\circ$
$\sin 250^\circ = \sin \left(270^\circ - 20^\circ\right) = - \cos 20^\circ$
$\cos 540^\circ = \cos \left(2 \times 180^\circ\right) = -1$
$\cos 290^\circ = \cos \left(270^\circ + 20^\circ\right) = \sin 20^\circ$
$\cos 430^\circ = \cos \left(450^\circ - 20^\circ\right) = \sin 20^\circ$
$\cos 1260^\circ = \cos \left(8 \pi - 180^\circ\right) = \cos 180^\circ = -1$
The given expression is: $\;\;$ $\dfrac{\sin 110^\circ \sin 250^\circ + \cos 540^\circ \cos 290^\circ \cos 430^\circ}{\cos^2 1260^\circ}$
$= \dfrac{\cos 20^\circ \times \left(- \cos 20^\circ\right) + \left(-1\right) \times \sin 20^\circ \times \sin 20^\circ}{\left(-1\right)^2}$
$= \dfrac{- \cos^2 20^\circ - \sin^2 20^\circ}{1}$
$= - \left(\cos^2 20^\circ + \sin^2 20^\circ\right) = -1$