Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 5 - Section 5.2 - Sum and Difference Formulas - Exercise Set - Page 669: 23


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 $2-\sqrt{3}$.
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