Answer
The exact value of $\tan \left( \frac{4\pi }{3}-\frac{\pi }{4} \right)$ is $2-\sqrt{3}$.
Work Step by Step
Use the difference formula of the tangent and evaluate the expression as,
$\tan \left( \frac{4\pi }{3}-\frac{\pi }{4} \right)=\frac{\tan \frac{4\pi }{3}-\tan \frac{\pi }{4}}{1+\tan \frac{4\pi }{3}\tan \frac{\pi }{4}}$
Substitute the values, $\tan \frac{4\pi }{3}=\sqrt{3}$ and $\tan \frac{\pi }{4}=1$.
$\begin{align}
& \tan \left( \frac{4\pi }{3}+\frac{\pi }{4} \right)=\frac{\sqrt{3}-1}{1+\left( \sqrt{3}\times 1 \right)} \\
& =\frac{\sqrt{3}-1}{\sqrt{3}+1}
\end{align}$
Further, rationalize the result and solve as,
$\begin{align}
& \tan \left( \frac{4\pi }{3}-\frac{\pi }{4} \right)=\frac{\sqrt{3}-1}{\sqrt{3}+1}\times \frac{\sqrt{3}-1}{\sqrt{3}-1} \\
& =\frac{{{\left( \sqrt{3}-1 \right)}^{2}}}{{{\left( \sqrt{3} \right)}^{2}}-{{1}^{2}}} \\
& =\frac{\left( 3+1-2\sqrt{3} \right)}{3-1} \\
& =\frac{4-2\sqrt{3}}{2}
\end{align}$
Thus,
$\begin{align}
& \tan \left( \frac{4\pi }{3}+\frac{\pi }{4} \right)=\frac{2\left( 2-\sqrt{3} \right)}{2} \\
& =2-\sqrt{3}
\end{align}$
Hence, the exact value of $\tan \left( \frac{4\pi }{3}+\frac{\pi }{4} \right)$ is equivalent to $2-\sqrt{3}$.
