MCQ
${\cos ^2}\left( {\frac{\pi }{6} + \theta } \right) - {\sin ^2}\left( {\frac{\pi }{6} - \theta } \right) = $
- ✓$\frac{1}{2}\cos 2\theta $
- B$0$
- C$ - \frac{1}{2}\cos 2\,\theta $
- D$\frac{1}{2}$
$ = \cos \left( {\frac{\pi }{6} + \theta + \frac{\pi }{6} - \theta } \right)\cos \left( {\frac{\pi }{6} + \theta - \frac{\pi }{6} + \theta } \right)$
$[ \because {\cos ^2}A - {\sin ^2}B = \cos (A + B)\cos (A - B)]$
$ = \cos \frac{{2\pi }}{6}\cos 2\theta = \frac{1}{2}\cos 2\theta $.
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