Answer
The result of $\cos \left( \frac{\pi }{6}+\alpha \right)\cos \left( \frac{\pi }{6}-\alpha \right)-\sin \left( \frac{\pi }{6}+\alpha \right)\sin \left( \frac{\pi }{6}-\alpha \right)$ is $\frac{1}{2}$.
Work Step by Step
Let us consider the given expression,
$\cos \left( \frac{\pi }{6}+\alpha \right)\cos \left( \frac{\pi }{6}-\alpha \right)-\sin \left( \frac{\pi }{6}+\alpha \right)\sin \left( \frac{\pi }{6}-\alpha \right)$
By using the trigonometric identity,
$\cos \left( \alpha +\beta \right)=\cos \alpha \cos \beta -\sin \alpha \sin \beta $
Now, the above expression can be further simplified as,
$\begin{align}
& \cos \left( \frac{\pi }{6}+\alpha \right)\cos \left( \frac{\pi }{6}-\alpha \right)-\sin \left( \frac{\pi }{6}+\alpha \right)\sin \left( \frac{\pi }{6}-\alpha \right)=\cos \left[ \left( \frac{\pi }{6}+\alpha \right)+\left( \frac{\pi }{6}-\alpha \right) \right] \\
& =\cos \left( \frac{\pi }{6}+\alpha +\frac{\pi }{6}-\alpha \right) \\
& =\cos \left( \frac{\pi }{6}+\frac{\pi }{6} \right) \\
& =\cos \left( \frac{2\pi }{6} \right)
\end{align}$
Use, $\cos \frac{\pi }{3}=\frac{1}{2}$.
$\begin{align}
& \cos \left( \frac{2\pi }{6} \right)=\cos \left( \frac{\pi }{3} \right) \\
& =\frac{1}{2}
\end{align}$
Thus, $\cos \left( \frac{\pi }{6}+\alpha \right)\cos \left( \frac{\pi }{6}-\alpha \right)-\sin \left( \frac{\pi }{6}+\alpha \right)\sin \left( \frac{\pi }{6}-\alpha \right)$ can be simplified as $\frac{1}{2}$.
Hence, the result of $\cos \left( \frac{\pi }{6}+\alpha \right)\cos \left( \frac{\pi }{6}-\alpha \right)-\sin \left( \frac{\pi }{6}+\alpha \right)\sin \left( \frac{\pi }{6}-\alpha \right)$ is $\frac{1}{2}$.
