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\begin{array}{l} Sans calculer  cos \frac{π}{12} et  sin \frac{π}{12}, démontrer que : \\ \frac{cos \frac{π}{12} + sin \frac{π}{12}}{cos \frac{π}{12} - sin \frac{π}{12}} = \sqrt{3} \end{array}

\begin{array}{l} \frac{cos \frac{π}{12} + sin \frac{π}{12}}{cos \frac{π}{12} - sin \frac{π}{12}} = \frac{(cos \frac{π}{12} + sin \frac{π}{12}) (cos \frac{π}{12} + sin \frac{π}{12})}{(cos \frac{π}{12} - sin \frac{π}{12}) (cos \frac{π}{12} + sin \frac{π}{12})} \\ = \frac{{(cos \frac{π}{12} + sin \frac{π}{12})}^{2}}{{cos}^{2} \frac{π}{12} - {sin}^{2} \frac{π}{12}} = \frac{{cos}^{2} \frac{π}{12} + {sin}^{2} \frac{π}{12} + 2sin \frac{π}{12} cos \frac{π}{12}}{cos (2 \times \frac{π}{12})} \\ \{car cos 2a = {cos}^{2} a - {sin}^{2} a\} \\ = \frac{1 + sin \frac{2 \times π}{12}}{cos \frac{π}{6}} \{car {cos}^{2} a + {sin}^{2} a = 1     et   sin 2a = 2sin a cos a\} \\ = \frac{1 + sin \frac{π}{6}}{cos \frac{π}{6}} = \frac{1 + \frac{1}{2}}{\frac{\sqrt{3}}{2}} = \frac{\frac{3}{2}}{\frac{\sqrt{3}}{2}} = \sqrt{3} \end{array}

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