## Precalculus (6th Edition) Blitzer

The given expression on the left side $\frac{\sec t-1}{t\sec t}$ can be further simplified by multiplying and dividing by $\cos t$. \begin{align} & \frac{\sec t-1}{t\sec t}=\frac{\sec t-1}{t\sec t}.\frac{\cos t}{\cos t} \\ & =\frac{\sec t.\cos t-\operatorname{cost}}{t\sec t.\cos t} \end{align} Now, the expression can be further simplified by using the reciprocal identity $\sec t=\frac{1}{\cos t}$ \begin{align} & \frac{\sec t.\cos t-\operatorname{cost}}{t\sec t.\cos t}=\frac{\frac{1}{\cos t}.\cos t-\cos t}{t\sec t.\cos t} \\ & =\frac{1-\cos t}{t} \end{align} Hence, the left side is equal to the right side $\frac{\sec t-1}{t\sec t}=\frac{1-\cos t}{t}$.