The value of cos

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

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

$=\cos ^{2} 48^{\circ}-\cos ^{2} 78^{\circ}$

$=(\cos 48-\cos 78)(\cos 48+\cos 78)$

$=\left(2 \sin 63^{\circ} \sin 15^{\circ}\right)\left(2 \cos 63^{\circ} \cos 15^{\circ}\right)$

Using identities :

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

$=\left(2 \sin 63^{\circ} \cos 63^{\circ}\right)\left(2 \sin 15^{\circ} \cos 15^{\circ}\right)$

$=\sin 126^{\circ} \sin 30^{\circ}$

$=\frac{1}{2} \sin 126^{\circ}=\frac{1}{2} \sin \left(90^{\circ}+36^{\circ}\right)=\frac{1}{2} \sin 36^{\circ}$ is the answer    $\left(\cdots \sin 30^{\circ}=\frac{1}{2}\right.$ and $\left.\sin \left(90^{\circ}+\theta\right)=\sin \theta\right)$