Answer
The required value is $\sqrt{3}+2$
Work Step by Step
The expression $\tan \frac{5\pi }{12}$ can be written as $\tan \left( \frac{2\pi }{12}+\frac{3\pi }{12} \right)$. Then use the identity:
$\tan \left( x+y \right)=\frac{\tan x+\tan y}{1-\tan x\tan y}$
Now, replacing the values in the above equation, we get:
$\begin{align}
& \tan \left( \frac{2\pi }{12}+\frac{3\pi }{12} \right)=\frac{\tan \frac{2\pi }{12}+\tan \frac{3\pi }{12}}{1-\tan \frac{2\pi }{12}\tan \frac{3\pi }{12}} \\
& =\frac{\tan \frac{\pi }{6}+\tan \frac{\pi }{4}}{1-\tan \frac{\pi }{6}.\tan \frac{\pi }{4}} \\
& =\frac{\frac{\sqrt{3}}{3}+1}{1-\frac{\sqrt{3}}{3}.1} \\
& =\frac{\frac{\sqrt{3}}{3}+1}{1-\frac{\sqrt{3}}{3}}.\frac{1+\frac{\sqrt{3}}{3}}{1+\frac{\sqrt{3}}{3}}
\end{align}$
The equation can be further simplified as:
$\begin{align}
& \frac{\frac{\sqrt{3}}{3}+1}{1-\frac{\sqrt{3}}{3}}.\frac{1+\frac{\sqrt{3}}{3}}{1+\frac{\sqrt{3}}{3}}=\frac{\frac{2\sqrt{3}}{3}+1+\frac{1}{3}}{1-\frac{3}{9}} \\
& =\frac{\frac{2\sqrt{3}}{3}+\frac{4}{3}}{1-\frac{1}{3}} \\
& =\frac{\frac{2\sqrt{3}}{3}+\frac{4}{3}}{\frac{2}{3}}
\end{align}$
On further simplifying, we get:
$\begin{align}
& \frac{\frac{2\sqrt{3}}{3}+\frac{4}{3}}{\frac{2}{3}}=\left( \frac{2\sqrt{3}}{3}+\frac{4}{3} \right).\frac{3}{2} \\
& =\sqrt{3}+2
\end{align}$
Hence, the required value is $\sqrt{3}+2$
