Prove that:

$\mathrm{LHS}=\sin ^{2} 42^{\circ}-\cos ^{2} 78^{\circ}$

$=\sin ^{2}\left(90^{\circ}-48^{\circ}\right)-\cos ^{2}\left(90^{\circ}-12^{\circ}\right)$

$=\cos ^{2} 48^{\circ}-\sin ^{2} 12^{\circ}$

$=\cos \left(48^{\circ}+12^{\circ}\right) \cos \left(48^{\circ}-12^{\circ}\right) \quad\left[\cos (\mathrm{A}+\mathrm{B}) \cos (\mathrm{A}-\mathrm{B})=\cos ^{2} \mathrm{~A}-\sin ^{2}\right]$

$=\cos 60^{\circ} \cos 36^{\circ}$

$=\frac{1}{2} \times \frac{\sqrt{5}+1}{4} \quad\left(\because \cos 36^{\circ}=\frac{\sqrt{5}+1}{4}\right)$

$=\frac{\sqrt{5}+1}{8}$

=RHS

Hence proved.