## Precalculus (6th Edition) Blitzer

To verify the given identity, $\sec x-\sec x{{\sin }^{2}}x=\cos x$ Recall Trigonometric Identities, \begin{align} & \sec x=\frac{1}{\cos x} \\ & {{\sin }^{2}}x+{{\cos }^{2}}x=1 \\ \end{align} Use the above identities and solve the left side of the given expression, \begin{align} & \sec x-\sec x{{\sin }^{2}}x=\sec x\left( 1-{{\sin }^{2}}x \right) \\ & =\left( \frac{1}{\cos x} \right)\left( {{\cos }^{2}}x \right) \\ & =\cos x \end{align} Hence, it is proved that the given identity holds true.