Question

How do I find the value of $\tan \left(\frac{\mathrm{\pi }}{12}\right)$.

Answer 1

Calculate the value of $\tan \left(\frac{\mathrm{\pi }}{12}\right)$ :

It is known that, $\mathbf{tan}\mathbf{(}\mathbf{A}\mathbf{-}\mathbf{B}\mathbf{)}\mathbf{=}\frac{\mathbf{tan}\mathbf{\left(}\mathbf{A}\mathbf{\right)}\mathbf{-}\mathbf{tan}\mathbf{\left(}\mathbf{B}\mathbf{\right)}}{\mathbf{1}\mathbf{+}\mathbf{tan}\mathbf{\left(}\mathbf{A}\mathbf{\right)}\mathbf{·}\mathbf{tan}\mathbf{\left(}\mathbf{B}\mathbf{\right)}}$.

Thus, $\tan \left(\frac{\mathrm{\pi }}{12}\right)=\tan \left(\frac{\mathrm{\pi }}{4}-\frac{\mathrm{\pi }}{6}\right)$

$$\begin{gathered} \Rightarrow \tan \left(\frac{\mathrm{\pi }}{12}\right)=\frac{\tan {\displaystyle \frac{\mathrm{\pi }}{4}}-\tan {\displaystyle \frac{\mathrm{\pi }}{6}}}{1+\left(\tan {\displaystyle \frac{\mathrm{\pi }}{4}}\right)\left(\tan {\displaystyle \frac{\mathrm{\pi }}{6}}\right)} \\ \Rightarrow \tan \left(\frac{\mathrm{\pi }}{12}\right)=\frac{1-{\displaystyle \frac{1}{\sqrt{3}}}}{1+\left(1\right)\left(\frac{1}{\sqrt{3}}\right)} \\ \Rightarrow \tan \left(\frac{\mathrm{\pi }}{12}\right)=\frac{{\displaystyle \frac{\sqrt{3}-1}{\sqrt{3}}}}{{\displaystyle \frac{\sqrt{3}+1}{\sqrt{3}}}} \\ \Rightarrow \tan \left(\frac{\mathrm{\pi }}{12}\right)=\frac{\sqrt{3}-1}{\sqrt{3}+1} \end{gathered}$$

Multiply both numerator and denominator by $\left(\sqrt{3}-1\right)$.

$$\begin{gathered} \Rightarrow \tan \left(\frac{\mathrm{\pi }}{12}\right)=\frac{\left(\sqrt{3}-1\right)\left(\sqrt{3}-1\right)}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)} \\ \Rightarrow \tan \left(\frac{\mathrm{\pi }}{12}\right)=\frac{{\left(\sqrt{3}\right)}^2-2\left(\sqrt{3}\right)\left(1\right)+{\left(1\right)}^2}{{\left(\sqrt{3}\right)}^2-{\left(1\right)}^2} \\ \Rightarrow \tan \left(\frac{\mathrm{\pi }}{12}\right)=\frac{3-2\sqrt{3}+1}{3-1} \\ \Rightarrow \tan \left(\frac{\mathrm{\pi }}{12}\right)=\frac{4-2\sqrt{3}}{2} \\ \Rightarrow \tan \left(\frac{\mathrm{\pi }}{12}\right)=2-\sqrt{3} \end{gathered}$$

Therefore, the value of $\tan \left(\frac{\mathrm{\pi }}{12}\right)$ is $\mathrm{(}\mathrm{2}\mathrm{-}\sqrt{\mathrm{3}}\mathrm{)}$.