Rezolvarea ta e corecta:
[tex]\dfrac{\sqrt 6-\sqrt 2}{\sqrt 6+ \sqrt 2} = \dfrac{\sqrt 3\cdot \sqrt 2 - \sqrt 2}{\sqrt 3\cdot \sqrt 2+\sqrt 2} = \dfrac{\sqrt 3 - 1}{\sqrt 3+1} = \dfrac{3+1-2\sqrt 3}{3-1} = \\ \\ = \dfrac{4-2\sqrt 3}{2} = 2-\sqrt 3[/tex]
Dar se făcea mai ușor așa:
[tex]\boxed{\mathrm{tg}\,\dfrac{x}{2} = \dfrac{\sin x}{1+\cos x}}\to \text{formula} \\ \\ \\\\\mathrm{tg}\,\dfrac{30^\circ}{2} = \dfrac{\sin 30^\circ}{1+\cos 30^\circ}= \dfrac{\dfrac{1}{2}}{1+\dfrac{\sqrt 3}{2}} = \dfrac{1}{2+\sqrt 3} = \dfrac{2-\sqrt 3}{4-3} = 2-\sqrt 3\\ \\ \Rightarrow \mathrm{tg}\, 15^\circ = 2-\sqrt 3[/tex]
