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: 21


The exact value of $\tan \left( \frac{\pi }{6}+\frac{\pi }{4} \right)$ is $2+\sqrt{3}$.

Work Step by Step

Use the sum formula of tangent and evaluate the expression as, $\tan \left( \frac{\pi }{6}+\frac{\pi }{4} \right)=\frac{\tan \frac{\pi }{6}+\tan \frac{\pi }{4}}{1-\tan \frac{\pi }{6}\tan \frac{\pi }{4}}$ Substitute the values, $\tan \frac{\pi }{6}=\frac{1}{\sqrt{3}}$ and $\tan \frac{\pi }{4}=1$. $\begin{align} & \tan \left( \frac{\pi }{6}+\frac{\pi }{4} \right)=\frac{\frac{1}{\sqrt{3}}+1}{1-\left( \frac{1}{\sqrt{3}}\times 1 \right)} \\ & =\frac{1+\sqrt{3}}{\sqrt{3}-1} \end{align}$ Further, rationalize the result and solve as, $\begin{align} & \tan \left( \frac{\pi }{6}+\frac{\pi }{4} \right)=-\left( \frac{1+\sqrt{3}}{1-\sqrt{3}} \right)\times \frac{1+\sqrt{3}}{1+\sqrt{3}} \\ & =-\frac{{{\left( 1+\sqrt{3} \right)}^{2}}}{{{1}^{2}}-{{\left( \sqrt{3} \right)}^{2}}} \\ & =-\frac{\left( 1+3+2\sqrt{3} \right)}{1-3} \\ & =\frac{4+2\sqrt{3}}{2} \end{align}$ Thus, $\begin{align} & \tan \left( \frac{\pi }{6}+\frac{\pi }{4} \right)=\frac{2\left( 2+\sqrt{3} \right)}{2} \\ & =2+\sqrt{3} \end{align}$ Hence, the exact value of $\tan \left( \frac{\pi }{6}+\frac{\pi }{4} \right)$ is equivalent to $2+\sqrt{3}$.
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