## Precalculus (6th Edition) Blitzer

The required solution is $\frac{\tan \theta +\tan \phi }{1-\tan \theta \tan \phi }$
We know the identity $\tan \left( \theta +\phi \right)$ is the sum of the tangent of the first angle and the tangent of second angle divided by 1 minus the product of both angles. The first step is to convert the tan in the form of sin and cos. As per the quotient identity of trigonometry $\tan \alpha =\frac{\sin \alpha }{\cos \alpha }$. Therefore, the expression can be simplified as: $\tan \left( \theta +\phi \right)=\frac{\sin \left( \theta +\phi \right)}{\cos \left( \theta +\phi \right)}$ Now, using the sum and difference identity of sin and cos and then multiplying both the numerator and denominator by $\frac{1}{\cos \theta \cos \phi }$. \begin{align} & \tan \left( \theta +\phi \right)=\frac{\sin \left( \theta +\phi \right)}{\cos \left( \theta +\phi \right)} \\ & =\frac{\sin \theta \cos \phi +\cos \theta \sin \phi }{\cos \theta \cos \phi -\sin \theta \sin \phi } \\ & =\frac{\sin \theta \cos \phi +\cos \theta \sin \phi }{\cos \theta \cos \phi -\sin \theta \sin \phi }\times \frac{\frac{1}{\cos \theta \cos \phi }}{\frac{1}{\cos \theta \cos \phi }} \end{align} Thus, the expression can be further simplified as: \begin{align} & \frac{\sin \theta \cos \phi +\cos \theta \sin \phi }{\cos \theta \cos \phi -\sin \theta \sin \phi }\times \frac{\frac{1}{\cos \theta \cos \phi }}{\frac{1}{\cos \theta \cos \phi }}=\frac{\frac{\sin \theta \cos \phi +\cos \theta \sin \phi }{\cos \theta \cos \phi }}{\frac{cos\theta \cos \phi -\sin \theta \sin \phi }{\cos \theta \cos \phi }} \\ & =\frac{\frac{\sin \theta \cos \phi }{\cos \theta \cos \phi }+\frac{\cos \theta \sin \phi }{\cos \theta \cos \phi }}{\frac{cos\theta \cos \phi }{\cos \theta \cos \phi }-\frac{\sin \theta \sin \phi }{\cos \theta \cos \phi }} \\ & =\frac{\frac{\sin \theta }{\cos \theta }.1+1.\frac{\sin \phi }{\cos \phi }}{1-\frac{\sin \theta }{\cos \theta }.\frac{\sin \phi }{\cos \phi }} \\ & =\frac{\tan \theta +\tan \phi }{1-\tan \theta \tan \phi } \end{align}