Question

O valor de tg 75° é igual a:

Answer 1

$tg~(a+b)=(tg~a+tg~b)/(1-tg~a*tg~b)$
________________________

$tg~75\º=tg~(30\º+45\º)\\tg~75\º=(tg~30\º+tg~45\º)/(1-tg~30\º*tg~45\º)\\tg~75\º=([\sqrt{3}/3]+1)/(1-[\sqrt{3}/3]*1)\\tg~75\º=([\sqrt{3}/3}]+1)/(1-[\sqrt{3}/3])\\tg~75\º=([\sqrt{3}+3]/3)/([3-\sqrt{3}]/3)\\tg~75\º=([\sqrt{3}+3]/3)*(3/[3-\sqrt{3}])\\tg~75\º=(\sqrt{3}+3)/(3-\sqrt{3})$

Racionalizando:

$tg~75\º=[(\sqrt{3}+3)(3+\sqrt{3})]/[(3-\sqrt{3})(3+\sqrt{3})]\\tg~75\º=[3\sqrt{3}+\sqrt{3}\sqrt{3}+3*3+3\sqrt{3}]/[3^{2}-\sqrt{3}^{2}]\\tg~75\º=[3\sqrt{3}+3+9+3\sqrt{3}]/[9-3]\\tg~75\º=[6\sqrt{3}+12]/6\\tg~75\º=6*[\sqrt{3}+2]/6\\tg~75\º=\sqrt{3}+2$

Answer 2

✅ Após resolver os cálculos, concluímos que o valor da tangente de 75° é:

         $\Large\displaystyle\text{ \begin{gathered}\boxed{\boxed{\:\:\:\bf \tan75^{\circ} = 2 + \sqrt{3}\:\:\:}}\end{gathered} }$

Seja o dado:

                   $\Large\displaystyle\begin{gathered} tg\:75^{\circ}\end{gathered}$

Sabendo que:

               $\Large\displaystyle\begin{gathered} tg\:75^{\circ} = \tan75^{\circ}\end{gathered}$            

Observe que a notação "tg" está em português e a notação "tan" está em inglês, porém são iguais. Vou utilizar a segunda notação, pelo fato da resposta ficar mais profissional no LaTeX.            

Sabendo que:

         $\Large\displaystyle\begin{gathered} \tan 75^{\circ} = \tan(30^{\circ} + 45^{\circ})\end{gathered}$

Para obter a tangente de 75° devemos utilizar a fórmula que calcula a tangente da soma de dois arcos que é:

      $\Large\displaystyle\begin{gathered} \tan(a + b) = \frac{\tan a + \tan b}{1 - \tan a\cdot\tan b}\end{gathered}$

Então, temos:

            $\Large\displaystyle\begin{gathered} \tan75 = \tan(30^{\circ} + 45^{\circ})\end{gathered}$

                           $\Large\displaystyle\text{ \begin{gathered} = \frac{\tan 30^{\circ} + \tan 45^{\circ}}{1 - \tan 30^{\circ} \cdot \tan 45^{\circ}}\end{gathered} }$

                            $\Large\displaystyle\text{ \begin{gathered} = \cfrac{\frac{\sqrt{3}}{3} + 1}{1 - \frac{\sqrt{3}}{3}\cdot1}\end{gathered} }$

                            $\Large\displaystyle\text{ \begin{gathered} = \cfrac{\frac{\sqrt{3} + 3}{3}}{1 - \frac{\sqrt{3}}{3}}\end{gathered} }$

                            $\Large\displaystyle\text{ \begin{gathered} = \cfrac{\frac{\sqrt{3} + 3}{3}}{\frac{3 - \sqrt{3}}{3}}\end{gathered} }$

                             $\Large\displaystyle\text{ \begin{gathered} = \frac{\sqrt{3} + 3}{\!\diagup\!\!\!\!3}\cdot\frac{\!\diagup\!\!\!\!3}{3 - \sqrt{3}}\end{gathered} }$

                             $\Large\displaystyle\text{ \begin{gathered} = \frac{\sqrt{3} + 3}{3 - \sqrt{3}}\end{gathered} }$

                             $\Large\displaystyle\text{ \begin{gathered} = \frac{\sqrt{3} + 3}{3 - \sqrt{3}}\cdot\frac{3 + \sqrt{3}}{3 + \sqrt{3}}\end{gathered} }$

                             $\Large\displaystyle\text{ \begin{gathered} = \frac{12 + 6\sqrt{3}}{9 - 3}\end{gathered} }$

                              $\Large\displaystyle\text{ \begin{gathered} = \frac{12 + 6\sqrt{3}}{6}\end{gathered} }$

                              $\Large\displaystyle\text{ \begin{gathered} = \frac{12}{6} + \frac{6\sqrt{3}}{6}\end{gathered} }$

                              $\Large\displaystyle\begin{gathered} = 2 + \sqrt{3}\end{gathered}$

✅ Portanto, o valor é:

                  $\Large\displaystyle\begin{gathered} \tan75^{\circ} = 2 + \sqrt{3}\end{gathered}$

Saiba mais:

1. https://brainly.com.br/tarefa/7599423
2. https://brainly.com.br/tarefa/547540
3. https://brainly.com.br/tarefa/4009100