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 - Concept and Vocabulary Check - Page 668: 8


The required solution is False.

Work Step by Step

We know the identity $\tan \left( \theta +\phi \right)$ is the sum of the tangent of the first angle and the tangent of the second angle divided by 1 minus the products of both angles. $\tan \left( x+y \right)=\frac{\tan x+\tan y}{1-\tan x\tan y}$ Then, replacing the values in the above equation, we get: $\begin{align} & \tan {{75}^{\circ }}=\tan \left( {{30}^{\circ }}+{{45}^{\circ }} \right)=\frac{\tan {{30}^{\circ }}+\tan {{45}^{\circ }}}{1-\tan {{30}^{\circ }}\tan {{45}^{\circ }}} \\ & =\frac{\frac{1}{\sqrt{3}}+1}{1-\frac{1}{\sqrt{3}}.1} \\ & =\frac{1+\frac{1}{\sqrt{3}}}{1-\frac{1}{\sqrt{3}}} \end{align}$ $\begin{align} & =\frac{1+\frac{1}{1.732}}{1-\frac{1}{1.732}} \\ & =\frac{1+0.577}{1-0.577} \\ & =\frac{1.577}{0.423} \\ & =3.728 \\ \end{align}$ $\begin{align} & \tan {{30}^{o}}+\tan {{45}^{o}}=\frac{1}{\sqrt{3}}+1 \\ & \text{ = 0}\text{.577+1} \\ & \text{ = 1}\text{.577} \\ \end{align}$ Thus, $\tan {{75}^{o}}\ne \tan {{30}^{o}}+\tan {{45}^{o}}$ Hence, the given statement is false.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.