## Precalculus (6th Edition) Blitzer

We have the expression on the left side $\frac{1-\tan x}{\sin x}$, which can be simplified by using the quotient identity $\tan x=\frac{\sin x}{\cos x}$ and the reciprocal identities $\frac{1}{\sin x}=\csc x$ and $\frac{1}{\cos x}=\sec x$ \begin{align} & \frac{1-\tan x}{\sin x}=\frac{1}{\sin x}-\frac{\tan x}{\sin x} \\ & =\frac{1}{\sin x}-\frac{\sin x}{\cos x}.\frac{1}{\sin x} \\ & =\frac{1}{\sin x}-\frac{1}{\cos x} \\ & =\csc x-\sec x \end{align} The expression on the left side is equal to the expression on the right side.