#### Answer

$4\rightarrow A$

#### Work Step by Step

$\cos ^{2}\alpha -\sin ^{2}\alpha =\cos 2\alpha \Rightarrow \cos ^{2}\dfrac {\pi }{6}-\sin ^{2}\dfrac {\pi }{6}=\cos \left( 2\times \dfrac {\pi }{6}\right) =\cos \dfrac {\pi }{3}=\dfrac {1}{2}\Rightarrow $
$4\rightarrow A$