## Precalculus (6th Edition) Blitzer

The given expression on the left side $\frac{1+\csc x}{\sec x}-\cot x$ can be further simplified by using the reciprocal identity $\csc x=\frac{1}{\sin x}$, $\sec x=\frac{1}{\cos x}$ and the quotient identity $\cot x=\frac{\cos x}{\sin x}$: \begin{align} & \frac{1+\csc x}{\sec x}-\cot x=\frac{1+\frac{1}{\sin x}}{\frac{1}{\cos x}}-\frac{\cos x}{\sin x} \\ & =\cos x.\left( 1+\frac{1}{\sin x} \right)-\frac{\cos x}{\sin x} \\ & =\cos x+\frac{\cos x}{\sin x}-\frac{\cos x}{\sin x} \\ & =\cos x \end{align} Hence, the left side is equal to the right side $\frac{1+\csc x}{\sec x}-\cot x=\cos x$