Resposta :
Observemos que:
[tex]\alpha (\circ)=\alpha (\text{rad})\cdot\dfrac{\pi}{180^{\circ}}[/tex]
Desta maneira, temos:
[tex]\dfrac{\pi}{12}=\dfrac{\pi}{12}\cdot\dfrac{180}{\pi}=15^{\circ}[/tex]
Então, podemos afirmar que:
[tex]\text{y}=\dfrac{1+\text{tg}~15^{\circ}}{1-\text{tg}~15^{\circ}}[/tex]
Como [tex]\text{tg}~15^{\circ}=2-\sqrt{3}[/tex], obtemos:
[tex]\text{y}=\dfrac{1+(2-\sqrt{3})}{1-(2-\sqrt{3})}[/tex]
[tex]\text{y}=\dfrac{3-\sqrt{3}}{-1+\sqrt{3}}[/tex]
Donde, temos:
[tex]\text{y}=\dfrac{3+3\sqrt{3}-\sqrt{3}-3}{1-3}[/tex]
[tex]\text{y}=\dfrac{-2\sqrt{3}}{-2}[/tex]
[tex]\text{y}=\sqrt{3}[/tex]