## Precalculus (6th Edition) Blitzer

Identity for $\tan \left( \theta -\phi \right)$ is expressed as $\frac{\tan \theta -\tan \phi }{1+\tan \theta \tan \phi }$.
From the difference formula of tangents, the tangent of the difference between two angles, say A and B, is expressed as, $\tan \left( A-B \right)=\frac{\tan A-\tan B}{1+\tan A\tan B}$ Thus, for angles $\theta$ and $\phi$ , $\tan \left( \theta -\phi \right)=\frac{\tan \theta -\tan \phi }{1+\tan \theta \tan \phi }$