## Precalculus (6th Edition) Blitzer

Let us consider the left side of the expression: \begin{align} & \sec \theta -\cos \theta =\frac{1}{\cos \theta }-\frac{\cos \theta }{1} \\ & =\frac{1-{{\cos }^{2}}\theta }{\cos \theta } \\ & =\frac{{{\sin }^{2}}\theta }{\cos \theta } \\ & =\frac{\sin \theta }{\cos \theta }\sin \theta \end{align} Finally, use the identities $\sec \theta =\frac{1}{\cos \theta },\ \tan \theta =\frac{\sin \theta }{\cos \theta }$. Therefore, $\sec \theta -\cos \theta =\tan \theta \sin \theta$ Thus, the left side of the expression is equal to right side.