${\cos }^2(\frac{\pi }{6}+\theta )-{\sin }^2(\frac{\pi }{6}-\theta )=$

${\cos }^2(\frac{\pi }{6}+\theta )-{\sin }^2(\frac{\pi }{6}-\theta )$

This is in the form of ${\cos }^{2}A-{\sin }^{2}B$

$=\cos (A+B)\cos (A-B)$

$\therefore A=30+\theta$

$B=30-\theta$

$=\cos (30+\theta +30-\theta ).\cos (30+\theta -30+\theta )$

$=\cos 60.\cos 2\theta$

$=\frac{1}{2}\cos 2\theta$