Precalculus (6th Edition) Blitzer

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

Chapter 5 - Review Exercises - Page 707: 16


The required value is $2-\sqrt{3}$

Work Step by Step

We know that the identity $\tan \left( \theta -\phi \right)$ is the difference between the tangent of the first angle and the tangent of the second angle divided by 1 plus the product of both angles. $\tan \left( \theta -\phi \right)=\frac{\tan \theta -\tan \phi }{1+\tan \theta \tan \left( \phi \right)}$ We use the above-mentioned identity where $\theta =\frac{4\pi }{3}$ and $\phi =\frac{\pi }{4}$ $\begin{align} & \tan \left( \frac{4\pi }{3}-\frac{\pi }{4} \right)=\frac{\tan \frac{4\pi }{3}-\tan \frac{\pi }{4}}{1+\tan \frac{4\pi }{3}.\tan \frac{\pi }{4}} \\ & =\frac{\sqrt{3}-1}{1+\sqrt{3}.1} \\ & =\frac{\sqrt{3}-1}{1+\sqrt{3}}.\frac{1-\sqrt{3}}{1-\sqrt{3}} \\ & =\frac{-{{\left( 1-\sqrt{3} \right)}^{2}}}{1-3} \end{align}$ And the above expression can be simplified by applying the algebraic identity ${{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}$ $\begin{align} & \frac{-{{\left( 1-\sqrt{3} \right)}^{2}}}{1-3}=\frac{-\left( 1-2.\sqrt{3}+3 \right)}{-2} \\ & =\frac{-\left( 4-2\sqrt{3} \right)}{-2} \\ & =\frac{-2\left( 2-\sqrt{3} \right)}{-2} \\ & =2-\sqrt{3} \end{align}$ Hence, the required value is $2-\sqrt{3}$
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