Respostas

  • Usuário do Brainly
2013-01-23T12:52:49-02:00

Observemos que:

 

\alpha (\circ)=\alpha (\text{rad})\cdot\dfrac{\pi}{180^{\circ}}

 

Desta maneira, temos:

 

\dfrac{\pi}{12}=\dfrac{\pi}{12}\cdot\dfrac{180}{\pi}=15^{\circ}

 

Então, podemos afirmar que:

 

\text{y}=\dfrac{1+\text{tg}~15^{\circ}}{1-\text{tg}~15^{\circ}}

 

Como \text{tg}~15^{\circ}=2-\sqrt{3}, obtemos:

 

\text{y}=\dfrac{1+(2-\sqrt{3})}{1-(2-\sqrt{3})}

 

\text{y}=\dfrac{3-\sqrt{3}}{-1+\sqrt{3}}

 

Donde, temos:

 

\text{y}=\dfrac{3+3\sqrt{3}-\sqrt{3}-3}{1-3}

 

\text{y}=\dfrac{-2\sqrt{3}}{-2}

 

\text{y}=\sqrt{3}