## Precalculus (6th Edition) Blitzer

The left side of the identity is as given below: $\frac{1}{\sin x\times \cos x}-\frac{\cos x}{\sin x}$ Simplifying the left side gives: \begin{align} & \frac{\sec x}{\sin x}-\frac{\cos x}{\sin x}=\frac{\sec x-\cos x}{\sin x} \\ & =\frac{\frac{1}{\cos x}-\cos x}{\sin x} \\ & =\frac{1-{{\cos }^{2}}x}{\sin x\times \cos x} \end{align} Simplifying further, \begin{align} & \frac{1-{{\cos }^{2}}x}{\sin x\times \cos x}=\frac{{{\sin }^{2}}x}{\sin x\times \cos x} \\ & =\frac{\sin x}{\cos x} \\ & =\tan x \end{align} Since, $\text{left side = right side}$ , we can say that the given identity is true. Thus, it is verified.