## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 5 - Section 5.2 - Sum and Difference Formulas - Exercise Set - Page 669: 56

#### Answer

The identity is $\frac{\tan \theta -\tan \phi }{1+\tan \theta \tan \phi }$.

#### Work Step by Step

We know the identity $\tan \left( \theta -\phi \right)$ is the difference between the tangent of the first angle and tangent of the second angle divided by 1 plus the product of both angles. By expressing the subtraction as an addition and using the additive identity $\tan \left( \theta +\phi \right)$, we obtain the result below. We recall that tan is odd -- that is, $\tan \left( -\theta \right)=-\tan \theta$. \begin{align} & \tan \left( \theta -\phi \right)=\tan \left( \theta +\left( -\phi \right) \right) \\ & =\frac{\tan \theta +\tan \left( -\phi \right)}{1-\tan \theta \tan \left( -\phi \right)} \\ & =\frac{\tan \theta -\tan \phi }{1+\tan \theta \tan \left( \phi \right)} \end{align} Hence, the identity for the given expression is $\frac{\tan \theta -\tan \phi }{1+\tan \theta \tan \phi }$.

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