## Precalculus (6th Edition) Blitzer

We have the expression on the left side: $\sec x-\cos x$, which can be simplified by using the reciprocal identity $\sec x=\frac{1}{\cos x}$. $\sec x-\cos x=\frac{1}{\cos x}-\cos x$ And the expression can be further simplified by multiplying and dividing the expression by $\cos x$. \begin{align} & \sec x-\cos x=\frac{1}{\cos x}-\cos x \\ & =\frac{1}{\cos x}-\frac{\cos x}{1}.\frac{\cos x}{\cos x} \\ & =\frac{1-{{\cos }^{2}}x}{\cos x} \end{align} We know the Pythagorean theorem ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$; therefore, ${{\sin }^{2}}\theta =1-{{\cos }^{2}}\theta$. Thus, we apply the theorem. And as per the quotient identity $\tan x=\frac{\sin x}{\cos x}$ \begin{align} & \frac{1-{{\cos }^{2}}x}{\cos x}=\frac{{{\sin }^{2}}x}{\cos x} \\ & =\frac{\sin x}{\cos x}.\sin x \\ & =\tan x.\sin x \end{align} Therefore, the left side is equal to the right side.