Question

calcule tg (75°) passo a passo

Answer 1

$tg\ 30 = \frac{ \sqrt{3} }{3} \ \ \ \ \ \ \ \ tg\ 45 = 1 \\ tg\ 75 =tg \ (30+45)= \frac{tg\ a + tg\ b}{1-tg\ a \cdot tg\ b} = \frac{tg\ 30 + tg\ 45}{1-tg\ 30 \\\cdot tg\ 45} =\frac{ \frac{ \sqrt{3}}{3} + 1}{1- \frac{\sqrt{3}}{3} \cdot 1}\\ \\tg\ 75 = \frac{ \frac{ \sqrt{3}+3}{3}}{1- \frac{\sqrt{3}}{3}} = \frac{ \frac{ \sqrt{3}+3}{3}}{ \frac{3-\sqrt{3}}{3}} = \frac{ \sqrt{3} +3}{3}\cdot \frac{3 }{3- \sqrt{3}} = \frac{ \sqrt{3}+3 }{3-\sqrt{3}}\cdot \frac{3+\sqrt{3}}{3+\sqrt{3}}$
$tg\ 75 = \frac{ \sqrt{3}+3 }{3-\sqrt{3}}\cdot \frac{3+\sqrt{3}}{3+\sqrt{3}} = \frac{(\sqrt{3})^2+2\cdot3\cdot\sqrt{3}+3^2}{3^2-(\sqrt{3})^2}= \frac{12+6\sqrt{3}}{9-3}= \frac{6\cdot(2+\sqrt{3})}{6}= 2+\sqrt{3}$

Answer 2

✅ Após resolver os cálculos, concluímos que o valor da soma 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:

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

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