## Precalculus (6th Edition) Blitzer

The exact value of $\tan \left( \frac{4\pi }{3}-\frac{\pi }{4} \right)$ is $2-\sqrt{3}$.
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 $2-\sqrt{3}$.